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