Question

We are giving the concept of A.M of $m^{th}$ power.Let $a,b>0$ and $a\ne b$ and let $m$ be a real number. Then
$\frac{a^m+b^m}{2}>{\left(\frac{a+b}{2}\right)}^m$ if $m\in R-[0,1]$
However, if $m\in (0,1)$ then $\frac{a^m+b^m}{2}<{\left(\frac{a+b}{2}\right)}^m$.
Obviously, if $m\in \{0,1\}$ then $\frac{a^m+b^m}{2}={\left(\frac{a+b}{2}\right)}^m$.
On the basis of the above information, answer the following questions:

If $x,y$ be positive real numbers, such that $x^2+y^2=8$ then max$(x+y)=$

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

Answer: (B)

$4$

From the above passage, we have the concept of A.M as for $a,b>0$ and

$m\in (0,1),$ $\frac{a^m+b^m}{2}<{\left(\frac{a+b}{2}\right)}^m$
Using this for $m=\frac{1}{2}$, we get

${\left(x^2\right)}^{\frac{1}{2}}+{\left(y^2\right)}^{\frac{1}{2}}\le {\left(\frac{x^2+y^2}{2}\right)}^{\frac{1}{2}}={\left(\frac{8}{2}\right)}^{\frac{1}{2}}$ ........ $(\because x^2+y^2=8)$

$\Rightarrow \frac{\left|x\right|+\left|y\right|}{2}\le 2$

$\Rightarrow \left|x\right|+\left|y\right|\le 4$
or $x+y\le 4$ ($\because x$ and $y$ are positive.)

Thus, max$(x+y)=4$