Problem on Inequality | ISI - MSQMS - B, 2018 | Problem 2a

Try this problem from ISI-MSQMS 2018 which involves the concept of Inequality.

INEQUALITY | ISI 2018| MSQMS | PART B | PROBLEM 2a


(a) Prove that if $x>0, y>0$ and $x+y=1,$ then $\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \geq 9$

Key Concepts


Algebra

Inequality

Numbers

Check The Answer


But Try the Problem First...

Answer: $xy \leq \frac{1}{4}$

ISI - MSQMS - B, 2018, Problem 2A

"INEQUALITIES: AN APPROACH THROUGH PROBLEMS BY BJ VENKATACHALA"

Try with Hints


We have to show that ,

$(1+\frac{1}{x})(1+\frac{1}{y}) \geq 9$

i.e $1+ \frac{1}{x} + \frac{1}{y} +\frac{1}{xy} \geq 9$

Since $x+y =1$

Therefore the above equation becomes $\frac{2}{xy} \geq 8$

ie $xy \leq \frac{1}{4}$

Now with this reduced form of the equation why don't you give it a try yourself,I am sure you can do it.

Applying AM $\geq$ GM on $x,y$

So you are just one step away from solving your problem,go on.............

Therefore, $\frac{x+y}{2} \geq (xy)^\frac{1}{2}$

$\Rightarrow \frac{1}{2} \geq (xy)^\frac{1}{2}$

Squaring both sides we get, $xy \leq \frac{1}{4}$

Hence the result follows.

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Inequality Problem From ISI - MSQMS - B, 2017 | Problem 3a

Try this problem from ISI-MSQMS 2017 which involves the concept of Inequality.

INEQUALITY | ISI 2017| MSQMS | PART B | PROBLEM 3a


(a) Prove that $a^{5}+b^{5}+c^{5}>a b c(a b+b c+c a),$ for all positive distinct values of $a, b, c$

Key Concepts


Algebra

Inequality

Numbers


Try with Hints


We know if we have $n$ numbers say $a_1,a_2,.....,a_n$ then AM $\geq$ GM implies

$\frac{a_1+a_2+....+a_n}{n} \geq (a_1.a_2........a_n)^\frac{1}{n}$

I assure you that the sum the can be done just by using this simple inequality,why don't you just give it a try?

Applying AM $\geq$ GM on $a^5,a^5,a^5,b^5,c^5$ we get

$3a^5+b^5+c^5 \geq 5a^3bc$

Similarly, $a^5+3b^5+c^5 \geq 5ab^3c$

$a^5+b^5+3c^5 \geq 5abc^3$

Adding the above three equations we get $a^5+b^5+c^5 \geq abc(a^2+b^2+c^2)$

So you have all the pieces of the jigsaw puzzle with you,and the puzzle is about to be completed,just try to place the remaining few pieces in its correct position

Now we have to show that $a^2+b^2+c^2 > ab+bc+ca$

Applying AM $\geq$ GM on $a^2,b^2$ we get,

$a^2+b^2 \geq 2ab$

Similarly,$b^2+c^2 \geq 2bc$

$a^2+c^2 \geq 2ca$

Adding the above three equations we get $a^2+b^2+c^2 \geq ab+bc+ca$

We are almost there ,so just try the last step yourself.

Therefore, $a^5+b^5+c^5 \geq abc(a^2+b^2+c^2) \geq abc(a^2+b^2+c^2)$

i.e, $a^{5}+b^{5}+c^{5}>a b c(a b+b c+c a)$

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Inequality Problem | ISI - MSQMS 2018 | Part B | Problem 4

Try this problem from ISI-MSQMS 2018 which involves the concept of Inequality and Combinatorics.

INEQUALITY | ISI 2018 | MSQMS | PART B | PROBLEM 4


Show that $\sqrt{C_{1}}+\sqrt{C_{2}}+\sqrt{C_{3}}+\ldots+\sqrt{C_{n}} \leq 2^{n-1}+\frac{n-1}{2}$ where
$C_k={n\choose k}$

Key Concepts


INEQUALITIES

COMBINATORICS


Try with Hints


Use Cauchy Schwarz Inequality $\left(\displaystyle\sum_{i} a_{i} b_{i}\right)^{2} \leq\left(\displaystyle\sum_{i} a_{i}^{2}\right)\left(\displaystyle\sum_{i} b_{i}^{2}\right)$

Apply Cauchy Schwarz Inquality in two sets of real numbers ($\sqrt C_1$,$\sqrt C_2$,.....,$\sqrt C_n$)and ($1$,$1$,$1$,......$1$)

($C_1+C_2+$........$+C_n$)($1+1+$......$+1$) $\geq $ ($\sqrt C_1+\sqrt C_2+.........+\sqrt C_n$)

($2^n-1$)$n \geq $ ($\sqrt C_1+\sqrt C_2+$..........$+\sqrt C_n$)$^2$

$\sqrt C_1+\sqrt C_2+$..........$+\sqrt C_n \leq \sqrt n\sqrt (2^n-1)$

The proof is still not done,why don't you try the remaining part yourself?

We know AM $\geq$ GM

i.e

For $n$ positive quantities $a_{1}, a_{2}, \dots, a_{n}$
$$
\frac{a_{1}+a_{2}+\ldots+a_{n}}{n} \geq \sqrt[n]{a_{1} a_{2} \cdot \cdot a_{n}}
$$
with equality if and only if $a_{1}=a_{2}=\ldots=a_{n}$

Now you have all the ingredients,why don't you cook it yourself? I firmly believe that you can cook a food tastier than mine.

$\frac{n+2^n-1}{2} \geq \sqrt n\sqrt {2^n-1}$

Thus,$\sqrt C_1+\sqrt C_2+$........$+\sqrt C_n \leq \frac {n+2^n-1}{2}$

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Mathematical Circles Inequality Problem

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]If \(a,b\) are positive reals such that \(a+b<2\) ,then prove that  $$ \displaystyle \frac {1}{1+a^2} + \frac {1}{1+b^2} \leq \frac {2}{1+ab} $$[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]Mathematical Circles[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Inequality involving AM-GM[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Mathematical Circles[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]Assume that the given inequality is true for \( a>0 , b>0 \) and \( a+b<2 \) . Then proceed .[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]In order to simplyfy the given inequality multiply both the sides by \( (1+a^2)(1+b^2)(1+ab) \) (as its a positive quantity and it is directly coming from   \( a>0 , b>0\)  ) .    [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]Come up with the simplest form of inequality  i.e. \( (a-b)^2 (1-ab) \geq 0 \) .[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]\( (a-b)^2 \geq 0 \) as \( a,b \in{R}\) . And to get \( (1-ab)>0 \) use the well known inequality for positive reals i.e. \( AM \geq GM \) and the still unused inequality i.e \( a+b <2 \) also . $$ \displaystyle a>0 , b>0 \Rightarrow \sqrt{ab}>0 \Rightarrow( 1+ \sqrt{ab})>0  \\  a>0 , b>0 , a+b <2 \Rightarrow  1 > \frac{a+b}{2} \geq \sqrt {ab} \\ \Rightarrow 1 > \sqrt{ab} \\ \Rightarrow ( 1 - \sqrt{ab}) >0 \\ \Rightarrow (1 - \sqrt{ab}) (1+ \sqrt{ab}) >0 \\ \Rightarrow (1 - ab)>0 $$[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="7" _address="0.1.0.7"]

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