Question

If $x>0,y>0$and $x^2+y^2=8[x\in R]$, then

• A. $x+y\ge 4$
• B. $x+y\le 4$
• C. $x+y=1$
• D. none of these

Answer 1

The correct option is B

$x+y\le 4$

Explanation for the correct answer:

To find the correct relation.

Given, $x^2+y^2=8$

We know, that the property of arithmetic mean says,

$\frac{\sqrt{x}+\sqrt{y}}{2}\le \sqrt{\left(\frac{x+y}{2}\right)}$

Now, we put $x=x^2\mathrm{and}y=y^2$

Then,

$$\begin{gathered} \frac{\sqrt{x^2}+\sqrt{y^2}}{2}\le \sqrt{\left(\frac{x^2+y^2}{2}\right)} \\ \Rightarrow \frac{x+y}{2}\le \sqrt{\left(\frac{x^2+y^2}{2}\right)} \\ \Rightarrow \frac{x+y}{2}\le \sqrt{\frac{8}{2}}\left[\because \mathrm{given},x^2+y^2=8\right] \\ \Rightarrow \frac{x+y}{2}\le 2 \\ \Rightarrow x+y\le 4 \end{gathered}$$

Hence, the correct option is (B) .