Question

If p and q are positive real numbers such that $p^2+q^2=1$, then the maximum value of $(p+q)$ is

Answer 1

Using A.M$\ge$G.M

$\Rightarrow \frac{p^2+q^2}{2}\ge pq$

$\Rightarrow pq\le \frac{1}{2}$ as $p^2+q^2=1$

We know that ${(p+q)}^2=p^2+q^2+2pq$

We have $p^2+q^2=1$

$\Rightarrow {(p+q)}^2-2pq=p^2+q^2$

$\Rightarrow {(p+q)}^2-2pq=1$

$\Rightarrow {(p+q)}^2\le 1+2pq$

$\Rightarrow {(p+q)}^2\le 1+2\times \frac{1}{2}$

$\Rightarrow {(p+q)}^2\le 1+1=2$

$\Rightarrow p+q\le \sqrt{2}$

Therefore,the maximum value of $(p+q)$=$\sqrt{2}$