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

First hint

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.

Second hint

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

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

Final Step

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.

## Other useful links

- https://www.cheenta.com/triple-integral-iit-jam-2016-question-15/
- https://www.youtube.com/watch?v=oUyHFKVB9IY