Try this problem from ISI-MSQMS 2018 which involves the concept of Inequality.
(a) Prove that if $x>0, y>0$ and $x+y=1,$ then $\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \geq 9$
Algebra
Inequality
Numbers
Answer: $xy \leq \frac{1}{4}$
ISI - MSQMS - B, 2018, Problem 2A
"INEQUALITIES: AN APPROACH THROUGH PROBLEMS BY BJ VENKATACHALA"
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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