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# Problem on Inequality | ISI – MSQMS – B, 2018 | Problem 2a

Try this problem from ISI MSQMS 2018 which involves the concept of Inequality. You can use the sequential hints provided to solve the problem.

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

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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