Question

If $a,b$ are natural numbers, prove that

$\sqrt{\frac{a^2+2b^2}{2}}>b$

Answer 1

$\sqrt{\frac{a^2+2b^2}{2}}>b$

Square on both sides

$\frac{a^2+2b^2}{2}>b^2$

$\Rightarrow a^2+2b^2>2b^2$

$\Rightarrow a^2>0$

$\Rightarrow a\in R$, $b$ is arbitrary

$\Rightarrow a,b\in R$.