Question

If $0\le x\le y$ and $(x+y{)}^2-(x-y{)}^2\ge 25$, what is the least possible value of $y$?

• A. $1.5$
• B. $2$
• C. $2.5$
• D. $3$

Answer 1

Answer: (C)

$2.5$

Given, $0$ $\le$ $x$ $\le$ $y$

Multiplying by $y$, we get

$0$ $\le$ $xy$ $\le$ $y^2$

Now, ${(x+y)}^2$ $-$ ${(x-y)}^2$ $\ge$ $25$

$\Rightarrow (x^2$ $+$ $y^2$ $+$ $2xy)$ $-$ $(x^2$ $+$ $y^2$ $-$ $2xy)$ $\ge$ $25$

$\Rightarrow x^2$ $+$ $y^2$ $+$ $2xy$ $-$ $x^2$ $-$ $y^2$ $+$ $2xy$ $\ge$ $25$

$\Rightarrow 4xy$ $\ge$ $25$

$\Rightarrow xy$ $\ge$ $\frac{25}{4}$

As $y^2$ $\ge$ $xy$ and $xy$ $\ge$ $\frac{25}{4}$,

Combining above two, we have

$y^2$ $\ge$ $xy$ $\ge$ $\frac{25}{4}$

$\Rightarrow y^2$ $\ge$ $\frac{25}{4}$

As $y>0$,

$\Rightarrow y$ $\ge$ $\sqrt{\frac{25}{4}}$

$\Rightarrow y$ $\ge$ $\frac{5}{2}$

$\Rightarrow y$ $\ge$ $2.5$

Therefore, the least possible value of $y$ is $2.5$.