An algebra problem by Anandmay Patel

$\begin{aligned} \frac {x^2+2}{\sqrt{x^2+1}} & = \sqrt{x^2+1} + \frac 1{\sqrt{x^2+1}} \end{aligned}$

Since both $\sqrt{x^2+1} > 0$ and $\dfrac 1{\sqrt{x^2+1}} > 0$ , we can apply the AM-GM inequality as follows:

$\begin{aligned} \sqrt{x^2+1} + \frac 1{\sqrt{x^2+1}} \ge 2 \sqrt{\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}} = 2 \end{aligned}$

Equality occurs when $\sqrt{x^2+1} = \dfrac 1{\sqrt{x^2+1}}$ $\implies x^2 + 1 = 1$ $\implies x = 0$ . Therefore, $\dfrac {x^2+2}{\sqrt{x^2+1}} > 2$ is true for all $x$ accept $0$ . That is it is $\boxed{\text{not always true}}$ .

$\text{We can proceed as follows:}$

$\dfrac{x^2+2}{\sqrt{x^2+1}}=\dfrac{x^2+1}{\sqrt{x^2+1}}+\dfrac1{\sqrt{x^2+1}}=\sqrt{x^2+1}+\dfrac1{\sqrt{x^2+1}}$

$\text{As the expression is of the form}$ $y+\dfrac1y$ , $\text{so it is always}$ $\textbf{equal to or greater than 2,and not always greater than 2}$ .

So the answer is $\textit{Not Always True}$ .

well, your process is all right, Anandmay. Just put x=0 and done.