The inequality is as follows:
$\frac{a^2+2}{\sqrt{a^2+1}}\ge2$
Ive tried squaring both sides and then making long division to simplify the left side, wich leaves me with a polynomial. I subtracted the 2 from both sides and tried to factor the left side, without success. What is a simpler way to do it?
Hint: write it as $ \;\cfrac{(a^2+1)+1}{\sqrt{a^2+1}} = \sqrt{a^2+1} + \cfrac{1}{\sqrt{a^2+1}}\ge2\,$.
Maybe AM-GM inequality, $(a^2+2)=(a^2+1)+1\geq 2\sqrt{a^2+1}$, so $\frac{a^2+2}{\sqrt{a^2+1}}\geq 2$
Observe the inequality is true when $a$ is replaced by $-a$. Thus we assume $a \ge 0$, and put $a = \tan \theta$, $0 \le \theta < \dfrac{\pi}{2}\implies$ $$\text{LHS} - \text{RHS}= \dfrac{\tan^2 \theta + 2}{\sec \theta}-2 = \dfrac{\sec^2 \theta +1}{\sec \theta} -2 = \cos \theta\left(\sec \theta - 1\right)^2 \ge 0.$$
It's true for all $a$. Set $x = a^2$, then you see $(x^2 + 2) > 0$ (square both sides). That is true for all real $x$.
a**2by $a^2$ andx**2by $x^2$. (I suppose that's what you wanted to write.) For some basic information about writing math at this site see e.g. here, here, here and here. – Martin Sleziak Jan 15 '17 at 07:35