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## Using auxiliary polynomials in ISI Entrance

Auxiliary polynomials are useful for solving complicated polynomial problems. This problem from ISI Entrance (and a Soviet Olympiad) is a good example.

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# Understand the problem

Let $a,b,c\ge-1$ be real numbers with $a^3+b^3+c^3=1$.
Prove that $a+b+c+a^2+b^2+c^2\le4$, and determine the cases of equality.
##### Source of the problem
Austria MO 2016. Final Round, Problem 4
Inequality
##### Suggested Book
Challenges and Thrills of Pre-College Mathematics

Do you really need a hint? Try it first!

The idea is that you have to capture the symmetry in the equations and correspondingly find it. Observe that the inequality $a+b+c+a^2+b^2+c^2\le4$ with the constraint $a^3+b^3+c^3=1$ can be written as  $(a^3 – a ^2 – a +1) + (b^3 – b ^2 – b +1) + (c^3 – c ^2 – c +1) \ge 0$ using the constraint.
Now, $(a^3 – a ^2 – a +1) + (b^3 – b ^2 – b +1) + (c^3 – c ^2 – c +1) \ge 0$ demands you to look into the polynomial  $P(x)=(x+1)(x-1)^2=x^3-x^2-x+1$. Thus, the problem reduces to show that if $a,b,c\ge-1$ are real numbers, then  $P(a)+P(b)+P(c)\ge0$.
What if we can show that individually if  $x\ge-1$ we always have $P(x)\ge0$? Then our problem will be solved right? We have $P(x)=(x+1)(x-1)^2=x^3-x^2-x+1$, observe that it automatically implies that if $x+1 \geq 0$ then we will have $P(x) \geq 0$.
Equality Cases: For equality we must have $P(a)=P(b)=P(c)=0$, and hence $a,b,c\in\{-1,+1\}$.
Hence equality holds if and only if one of the three variables is $-1$ and the other two are $+1$. QED

# Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

## Triangle Problem | PRMO-2018 | Problem No-24

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## What is Parity in Mathematics ? 🧐

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## Good numbers Problem | PRMO-2018 | Question 22

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## Polynomial Problem | PRMO-2018 | Question 30

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## Digits Problem | PRMO – 2018 | Question 19

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## Chocolates Problem | PRMO – 2018 | Problem No. – 28

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## Trigonometry | PRMO-2018 | Problem No-14

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Categories

# Understand the problem

Suppose that real numbers $x,y$ and $z$ satisfy the following equations:

\begin{align*} x+\frac{y}{z} &=2,\\ y+\frac{z}{x} &=2,\\ z+\frac{x}{y} &=2. \end{align*}
Show that $s=x+y+z$ must be equal to $3$ or $7$.

Note: It is not required to show the existence of such numbers $x,y,z$.

##### Source of the problem

Germany MO 2019, Problem 6

##### Topic
Algebra, Simultaneous Equations
6/10
##### Suggested Book
Challenges and Thrills of Pre College Mathematics

Do you really need a hint? Try it first!

Observe that x = y = z = 1 gives a valid solution of the set of equations. In this case s = x+y+z = 3.

Now, observe one thing that this set of equations is symmetric in (x,y,z).

Observe that we are required to comment on (x+y+z).

Rewriting the equations as:

$xz+y = 2z, \qquad (1)$
$xy + z = 2x, \qquad (2)$
$yz + x = 2y \qquad (3)$
and then summing gives us that $x+y+z = xy + yz + zx = s.$

Our aim will be to reduce all the equations into a single variable. ( maybe a polynomial ).

Let’s consider the case, where all of x,y,z is not 1.

From now on we consider $x,y,z \neq 1$. This also gives $x \neq y \neq z \neq x$

Solving the first expression

$x=\frac{2z-y}{z}$

then plugging this into the second two gives:

$y+\frac{z^2}{2z-y}=2 \Rightarrow (2z-y)y+z^2=2(2z-y)$$z+\frac{2z-y}{yz}=2 \Rightarrow yz^2+2z-y=2yz \Rightarrow y=-\frac{2z}{z^2-2z-1}$

as z is not equal to 1.

Plugging the latter into the former and simplifying gives:

$\frac{z^2 (z^4-8z^3+14z^2-7)}{(z^2-2z-1)^2}=0 \Rightarrow z^4-8z^3+14z^2-7=0$

Plugging the latter into the former and simplifying gives:

$\frac{z^2 (z^4-8z^3+14z^2-7)}{(z^2-2z-1)^2}=0 \Rightarrow z^4-8z^3+14z^2-7=0$

Now, observe that we already know z = 1 is a solution. This gives rise to

$0=z^4-8z^3+14z^2-7=(z-1)(z^3-7z^2+7z+7) \Rightarrow z^3-7z^2+7z+7=0$

Observe that the polynomial we have got in terms of z is also satisfied by x,y,z as the equations are symmetric in x,y,z.

Hence we can claim that $t^3 – 7t^2 + 7t + 7 = 0$ has three solutions x,y,z.

Hence, $t^3 – 7t^2 + 7t + 7 = (t-x)(t-y)(t-z)$.

Therefore, by Vieta’s formula, x+y+z = 7.

QED.

# Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year.

Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

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Try this beautiful Problem on Trigonometry from PRMO -2018.You may use sequential hints to solve the problem.

## What is Parity in Mathematics ? 🧐

Parity in Mathematics is a term which we use to express if a given integer is even or odd. It is basically depend on the remainder when we divide a number by 2. Parity can be divided into two categories – 1. Even Parity 2. Odd Parity Even Parity : If we...

## Value of Sum | PRMO – 2018 | Question 16

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## Chessboard Problem | PRMO-2018 | Problem No-26

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## Measure of Angle | PRMO-2018 | Problem No-29

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## Good numbers Problem | PRMO-2018 | Question 22

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## Polynomial Problem | PRMO-2018 | Question 30

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## Digits Problem | PRMO – 2018 | Question 19

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## Chocolates Problem | PRMO – 2018 | Problem No. – 28

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## Trigonometry | PRMO-2018 | Problem No-14

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Categories

# Understand the problem

Given the polynomial $P(x)=(x^2-7x+6)^{2n}+13$ where $n$ is a positive integer. Prove that $P(x)$ can’t be written as a product of $n+1$ non-constant polynomials with integer coefficients.
##### Source of the problem
Vietnam MO 2014 Problem 2
##### Topic
Polynomial, Algebra
7/10
##### Suggested Book
Excursion in Mathematics by Bhaskaracharya Prathistan

Do you really need a hint? Try it first!

The idea is that if a polynomial can be factorized, then we must check if a polynomial has real roots or not.

The first observation is that the given polynomial is always > 0 as it is of the form $(f(x))^2 + 13$.

Hence, it has no real roots.

So, if the polynomial can be factorized, then obviously the factors will not be linear.

An odd degree polynomial has atleast one real root. Hence, none of the factors will be odd degree.

So, we have seen the factors will not be linear. Hence, let’s investigate the factors and their properties if the polynomial can be factorized.

Let us approach the method of contradiction.

Assume that the polynomial can be factorized into n+1 factors with integer coefficients.

Now, we know that the factors must have atleast degree 2, in fact only even degrees i.e. 2, 4, 6, …

Now, let’s see if all the factors have a degree more than 2, then the degree of the whole polynomial must be $\geq 4n + 4$, but the degree of the polynomial is 4n.

So, two of the factors must be of degree 2.

Hence there exist $a,b, c, d$ such that $(x^2+ax+b)(x^2+cx+d) \mid P(x)$.

Note that $f(x)=x^2+ax+b > 0$ and $g(x)=x^2+cx+d > 0$ for all $x \in \mathbb R$, because they cannot have a real root.

Now, we have to make use of the idea that the coefficients are integers. So, some idea of divisibility must come in.

The given polynomial is $P(x)=(x^2-7x+6)^{2n}+13$ .

Try to find out some values of x, for which P(x) is easily determined without n.

See, x = 1 and 6 works as $x^2 – 7x + 6 = (x-1)(x-6)$.

So, we saw that x = 1 and x = 6 are good values and we get that P(1) = P(6) = 13.

Also, we know that f(1).g(1) | P(1) = 13. So, one of them must be 1.

Say, f(1) = 1 = 1 + a + b.

Also, f(6) = 36 + 6a – a = 36 + 5a.

Now, f(6) | P(6) = 13. So, f(6) = 36 + 5a = 1 or 13.

This gives rise to a single case of a = -7; where $f(x) = x^2 -7x + 7$.

But f(x) has real roots implying that P(x) has real roots. Hence contradiction.

QED.

# Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year.

Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

## Triangle Problem | PRMO-2018 | Problem No-24

Try this beautiful Problem on Trigonometry from PRMO -2018.You may use sequential hints to solve the problem.

## What is Parity in Mathematics ? 🧐

Parity in Mathematics is a term which we use to express if a given integer is even or odd. It is basically depend on the remainder when we divide a number by 2. Parity can be divided into two categories – 1. Even Parity 2. Odd Parity Even Parity : If we...

## Value of Sum | PRMO – 2018 | Question 16

Try this Integer Problem from Number theory from PRMO 2018, Question 16 You may use sequential hints to solve the problem.

## Chessboard Problem | PRMO-2018 | Problem No-26

Try this beautiful Problem on Trigonometry from PRMO -2018.You may use sequential hints to solve the problem.

## Measure of Angle | PRMO-2018 | Problem No-29

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## Good numbers Problem | PRMO-2018 | Question 22

Try this good numbers Problem from Number theory from PRMO 2018, Question 22 You may use sequential hints to solve the problem.

## Polynomial Problem | PRMO-2018 | Question 30

Try this Integer Problem from Number theory from PRMO 2018, Question 30 You may use sequential hints to solve the problem.

## Digits Problem | PRMO – 2018 | Question 19

Try this Integer Problem from Number theory from PRMO 2018, Question 19 You may use sequential hints to solve the problem.

## Chocolates Problem | PRMO – 2018 | Problem No. – 28

Try this beautiful Problem on Combinatorics from PRMO -2018.You may use sequential hints to solve the problem.

## Trigonometry | PRMO-2018 | Problem No-14

Try this beautiful Problem on Trigonometry from PRMO -2018.You may use sequential hints to solve the problem.

Categories

# Understand the problem

Find all the real Polynomials P(x) such that it satisfies the functional equation:

$P(2P(x)) = 2P(P(x)) + P(x)^{2} \forall real x$.

Unknown

##### Topic
Functional Equation, Polynomials
7/10
##### Suggested Book
Excursion in Mathematics

Challenges and Thrills in Pre College Mathematics

Do you really need a hint? Try it first!

Well, it is really good that the information polynomial is given! You should use that.

What is the first thing that you check in a Polynomial Identity?

Degree! Yes, check whether the degree of the Polynomial on both the LHS and RHS are the same or not. Yes, they are both the same $n^2$.  But did you observe something fishy?

Now rewrite the equation as $P(2P(x)) - 2P(P(x)) = P(x)^{2}$.

Do the Degree trick now… You see it right?

Yes, on the left it is $n^2$ and on the RHS it is $2n$.

So, there are two cases now… Figure them out!

Case 1: $2n = n^2$… i.e. P(x) is either a quadratic or a constant function.

Case 2:  $P(2P(x)) - 2P(P(x))$ has coefficient zero till $x^2n$.

We will study case 1 now.

Case 1: $2n = n^2$… i.e. P(x) is either a quadratic or a constant function.

$P(2P(x)) - 2P(P(x)) = P(x)^{2}$ = $P(2y) - 2P(y) = y^{2}$ where $y = P(x)$.

Now, expand using $P(x) = ax^2 + bx +c$, it gives $2ay^2 -c = y^2$… Now find out all such polynomials satisfying this property.

For e.g. $\frac{x^2}{2}$ is a solution.

If P(x) is constant, prove that $P(x) = 0 / \frac{-1}{2}$.

Case 2: $P(2y) - 2P(y) = y^{2}$.

Assume a general form of P(x) = $latex$and show that P(x) must be quadratic or lesser degree by comparing coefficients as you have a quadratic on RHS and n degree polynomial of the LHS.

• Always Compare the Degree of Polynomials in identities like this. It provides a lot of information.
• Compare the coefficients of Polynomials on both sides to equalize the coefficient on both sides.
• Find all polynomials $P(2P(x)) - 8P(P(x)) = P(x)^{2}$.
• Find all polynomials $P(cP(x)) – d P(P(x)) = P(x)^{2}$ depending on the values of c and d.

# Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year.

Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

## Triangle Problem | PRMO-2018 | Problem No-24

Try this beautiful Problem on Trigonometry from PRMO -2018.You may use sequential hints to solve the problem.

## What is Parity in Mathematics ? 🧐

Parity in Mathematics is a term which we use to express if a given integer is even or odd. It is basically depend on the remainder when we divide a number by 2. Parity can be divided into two categories – 1. Even Parity 2. Odd Parity Even Parity : If we...

## Value of Sum | PRMO – 2018 | Question 16

Try this Integer Problem from Number theory from PRMO 2018, Question 16 You may use sequential hints to solve the problem.

## Chessboard Problem | PRMO-2018 | Problem No-26

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## Measure of Angle | PRMO-2018 | Problem No-29

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## Good numbers Problem | PRMO-2018 | Question 22

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## Polynomial Problem | PRMO-2018 | Question 30

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## Digits Problem | PRMO – 2018 | Question 19

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## Chocolates Problem | PRMO – 2018 | Problem No. – 28

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## Trigonometry | PRMO-2018 | Problem No-14

Try this beautiful Problem on Trigonometry from PRMO -2018.You may use sequential hints to solve the problem.

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# Understand the problem

Find all the polynomials $P(x)$ of a degree $\leq n$ with real non-negative coefficients such that $P(x) \cdot P(\frac{1}{x}) \leq [P(1)]^2$ , $\forall x>0$.
##### Source of the problem
Albanian BMO TST 2009
Algebra
Easy
##### Suggested Book
An Excursion in Mathematics

Do you really need a hint? Try it first!

This problem is all about non-negative real numbers. The first thing that should come to your mind is “standard inequalities!”.
Write $P(x)=\Sigma a_kx^k$. Using the Cauchy-Schwarz inequality, show that $P(x)P(1/x)\ge (P(1))^2$.
Note that hint 2 along with the hypothesis in the problem implies that $P(x)P(1/x)=(P(1))^2$. Hence equality holds in hint 2.
As equality holds in CS, it means that for all $k$ satisfying $a_k\neq 0$, $\frac{a_kx^k}{a_kx^{-k}=x^{2k}$ is a constant. This is absurd, hence there can be at most one such $k$. Hence, only monomials can satisfy the given inequality.

# Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year.

Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

## Triangle Problem | PRMO-2018 | Problem No-24

Try this beautiful Problem on Trigonometry from PRMO -2018.You may use sequential hints to solve the problem.

## What is Parity in Mathematics ? 🧐

Parity in Mathematics is a term which we use to express if a given integer is even or odd. It is basically depend on the remainder when we divide a number by 2. Parity can be divided into two categories – 1. Even Parity 2. Odd Parity Even Parity : If we...

## Value of Sum | PRMO – 2018 | Question 16

Try this Integer Problem from Number theory from PRMO 2018, Question 16 You may use sequential hints to solve the problem.

## Chessboard Problem | PRMO-2018 | Problem No-26

Try this beautiful Problem on Trigonometry from PRMO -2018.You may use sequential hints to solve the problem.

## Measure of Angle | PRMO-2018 | Problem No-29

Try this beautiful Problem on Trigonometry from PRMO -2018.You may use sequential hints to solve the problem.

## Good numbers Problem | PRMO-2018 | Question 22

Try this good numbers Problem from Number theory from PRMO 2018, Question 22 You may use sequential hints to solve the problem.

## Polynomial Problem | PRMO-2018 | Question 30

Try this Integer Problem from Number theory from PRMO 2018, Question 30 You may use sequential hints to solve the problem.

## Digits Problem | PRMO – 2018 | Question 19

Try this Integer Problem from Number theory from PRMO 2018, Question 19 You may use sequential hints to solve the problem.

## Chocolates Problem | PRMO – 2018 | Problem No. – 28

Try this beautiful Problem on Combinatorics from PRMO -2018.You may use sequential hints to solve the problem.

## Trigonometry | PRMO-2018 | Problem No-14

Try this beautiful Problem on Trigonometry from PRMO -2018.You may use sequential hints to solve the problem.

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## RMO 2017 Problem 3 – Roots of a Polynomial

Here is a video post that discusses the roots of a polynomial problem from RMO 2017 problem 3. Watch, learn and enjoy the video.

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## TIFR 2013 Problem 38 Solution -Eigenvalue of differentiation

TIFR 2013 Problem 38 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Linear Algebra Done Right by Sheldon Axler. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem Type:True/False?

Let (V) be the vector space of polynomials with real coefficients in variable (t) of degree ( \le 9). Let (D:V\to V) be the linear operator defined by (D(f):=\frac{df}{dt}). Then (0) is an eigenvalue of (D).

Hint:

If 0 were an eigenvalue, what would be its eigenvector?

Discussion:

There are several ways to do this. One possible way is to find out the matrix representation of (D) with respect to standard basis ( {1,t,t^2,…,t^n})( or any other basis) and observe that it is a (strictly) upper triangular matrix with all diagonal entries 0 and therefore the determinant of (D) is 0. This implies that D is not injective, so there is some nonzero vector to which when (D) is applied gives the zero vector. Therefore, (D) has 0 eigenvalue.

Another way to do this is by observing that (D(1)=0(1)), therefore 0 is an eigenvalue of (D) with 1 as an eigenvector.

## HELPDESK

• What is this topic: Linear Algebra
• What are some of the associated concept: eigenvector,eigenvalue
• Book Suggestions: Linear Algebra done Right by Sheldon Axler

Categories

## Sum of polynomials | Tomato subjective 173

Try this beautiful problem from TOMATO Subjective Problem no. 173 based on the Sum of Polynomials.

Problem : Sum of polynomials

Let ${{P_1},{P_2},...{P_n}}$ be polynomials in ${x}$, each having all integer coefficients, such that ${{P_1}={{P_1}^{2}+{P_2}^{2}+...+{P_n}^{2}}}$. Assume that ${P_1}$ is not the zero polynomial. Show that ${{P_1}=1}$ and ${{P_2}={P_3}=...={P_n}=0}$

Solution :

As ${P_1},{P_2},...{P_n}$ are integer coefficient polynomials so gives integer values at integer points.

Now as ${P_1}$ is not zero polynomial

${\displaystyle{P_1}(x)>0}$ for some $\displaystyle{x \in Z}$

Then ${\displaystyle{P_1}(x)\ge{1}}$ or $P_1(x) \le -1$ as ${\displaystyle{P_1}(x)}$=integer

$\Rightarrow (P_1(x))^2 \ge P_1(x)$ or $0 \ge P_1(x) -(P_1(x))^2$

But it is given that ${{P_1}={{P_1}^{2}+{P_2}^{2}+...+{P_n}^{2}}}$

This implies $(P_2 (x))^2+...+(P_n (x) )^2 \le 0$. This is only possible if  $(P_1(x))^2 = ... = (P_n(x))^2 = 0$

Hence the values of x for which $P_1 (x)$ is non-zero, $P_2(x) , ... , P_n(x)$ are all zero. The values of x for which $P_1(x) = 0$,  we have  $0=0+(P_2 (x))^{2}+...+(P_n(x))^2$ implying each is zero.

Therefore $P_2(x) = ... = P_n(x) = 0$.

Finally $P_1(x) = (P_1(x))^2$ implies $P_1(x) = 0 \text{or} P_1(x) = 1$. Since $P_1(x) \neq 0$ hence it is 1.

(Proved)

Categories

## A Cauchy Schwarz Problem

Cauchy Schwarz Problem: Let $P(x)$ be a polynomial with non-negative coefficients.Prove that if $P(x)\cdot P(\frac{1}{x})\ge1$ for $x=1$,then the same inequality holds for each $\mathbb{R^+} x$.

Discussion: Cauchy Schwarz’s Inequality: Suppose for real numbers (\ a_{i},b_{i}), where (\ i\in{1,2,\dots,n}) we can say that $${\sum_{i=1}^{n}a_{i}^2}{\sum_{i=1}^{n}b_{i}^2}=\sum_{i=1}^{n}{a_{i}b_{i}}^2$$.

Titu’s Lemma: Let (\ a_{i},b_{i}\in{\mathbb{R}}) and let (\ a_{i},b_{i}>0) for (\ i\in{1,2,\dots,n})

$$\sum_{i=1}^{n}\frac{a_{i}^2}{b_{i}}\ge\frac{{\sum_{i=1}^{n}a_{i}}^2}{\sum_{i=1}^{n}b_{i}}$$

Proof of Cauchy Schwarz’s Inequality: We can write (\sum_{i=1}{n}a_{i}^2=\sum_{i=1}^{n}\frac{a_{i}^2 b_{i}^2}{b_{i}^2}\ge\frac{{\sum_{i=1}^{n}a_{i}b_{i}}^2}{\sum_{i=1}^{n}b_{i}^2})    (using Titu’s lemma)

(=>{\sum_{i=1}^{n}a_{i}^2}{\sum_{i=1}^{n}b_{i}^2}\ge {\sum_{i=1}^{n}{a_{i}b_{i}}^2)

Solution: Let $P(x)=a_{n}x^n+a_{n-1}x^{n-1}+\dots+a_{1}x+a_{0}$,then $P(\frac{1}{x})=a_{n}\frac{1}{x^n}+a_{n-1}\frac{1}{x^{n-1}}+\dots+a_{1}\frac{1}{x}+a_{0}$

now $P(x)\cdot P(\frac{1}{x})=\{a_{n}x^n+a_{n-1}x^{n-1}+\dots+a_{1}x+a_{0}\}\{a_{n}\frac{1}{x^n}+a_{n-1}\frac{1}{x^{n-1}}+\dots+a_{1}\frac{1}{x}+a_{0}\}=\{\{\sqrt{a_{n}x^n}\}^2+\{\sqrt{a_{n-1}x^{n-1}}\}^2+\dots+\{\sqrt{a_{1}x}\}^2+\{\sqrt{a_{0}}\}^2\} \{\{\sqrt{a_{n}\frac{1}{x^n}}\}^2+\{\sqrt{a_{n-1}\frac{1}{x^{n-1}}}\}^2+\dots+\{\sqrt{a_{1}\frac{1}{x}}\}^2+\{\sqrt{a_{0}}\}^2\} \ge\{a_{n}+a_{n-1}+\dots+a_{1}+a_{0}\}^2$
(Cauchy Schwarz’s Inequality)

now it is given that $P(x)\cdot P(\frac{1}{x})\ge1$ for $x=1$,so $P(1)^2\ge 1=>\{a_{n}+a_{n-1}+\dots+a_{1}+a_{0}\}^2\ge1$

therefore,$P(x)\cdot P(\frac{1}{x})\ge1$ $\forall x\in{\mathbb{R^+}}$