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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?

4 Answers 4

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 $x$.

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    a2 means a-squared, and x2 means x -squared. . . . – 2017-01-15
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    I replace `a**2` by $a^2$ and `x**2` by $x^2$. (I suppose that's what you [wanted to write](http://math.stackexchange.com/revisions/be049ea9-6997-4fea-9224-57683e988201/view-source).) For some basic information about writing math at this site see e.g. [here](/help/notation), [here](//meta.math.stackexchange.com/q/5020), [here](//meta.stackexchange.com/a/70559) and [here](//meta.math.stackexchange.com/q/1773). – 2017-01-15
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    But I do not see how this is related to the result wanted by the OP. After the substitution $x=a^2$, it is $\frac{x+2}{\sqrt{x+1}} \ge 2$. Did you perhaps want to say $y=a^2+1$ and somehow get to $y+1\ge2\sqrt y$? – 2017-01-15
3

Hint: write it as $ \;\cfrac{(a^2+1)+1}{\sqrt{a^2+1}} = \sqrt{a^2+1} + \cfrac{1}{\sqrt{a^2+1}}\ge2\,$.

2

Maybe AM-GM inequality, $(a^2+2)=(a^2+1)+1\geq 2\sqrt{a^2+1}$, so $\frac{a^2+2}{\sqrt{a^2+1}}\geq 2$

2

Observe the inequality is true when $a$ is replaced by $-a$. Thus we assume $a \ge 0$, and put $a = \tan \theta$, $0 \le \theta < \dfrac{\pi}{2}\implies$ $$\text{LHS} - \text{RHS}= \dfrac{\tan^2 \theta + 2}{\sec \theta}-2 = \dfrac{\sec^2 \theta +1}{\sec \theta} -2 = \cos \theta\left(\sec \theta - 1\right)^2 \ge 0.$$