# Proving with the Am Gm inequality [closed]

If a and b are any real numbers, prove $\frac{4a^2b}{a^4+b^4+1} \le \sqrt{2}$

So far it's obvious that if b is negative, the inequality holds. However I'm not sure how to change the equation into an am gm inequality which proves this inequality when b is positive.

## closed as off-topic by Arnaldo, Zain Patel, Daniel W. Farlow, Shailesh, JMPMay 6 '17 at 2:43

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So, $$\frac{a^4}{2}+\frac{a^4}{2}+b^4+1 \geq 4\sqrt[4]{\frac{a^8b^4}{4}} = 2\sqrt{2}a^2b.$$ Rearranging, we get the result: $$\frac{4a^2b}{a^4+b^4+1}\leq \sqrt{2}.$$
If $a=0\,$ or $b \le 0$, then $\text{LHS} \le 0 < \sqrt{2}$.
Next, suppose $a \ne 0$ and $b>0.\;\,$Then \begin{align*} \frac{a^4+b^4+1}{4a^2b} &=\frac{a^4+ (b^4+1)}{4a^2b}\\[6pt] &\ge\frac{a^4+ 2b^2}{4a^2b}&&\text{[by AM-GM]}\\[6pt] &\ge\frac{2\sqrt{(a^4)(2b^2)}}{4a^2b}&&\text{[by AM-GM]}\\[6pt] &=\frac{2a^2b\sqrt{2}}{4a^2b}\\[6pt] &=\frac{\sqrt{2}}{2}\\[6pt] &=\frac{1}{\sqrt{2}}\\[20pt] \text{Then}\;\;\frac{a^4+b^4+1}{4a^2b}&\ge\frac{1}{\sqrt{2}}\\[6pt] \implies\;\frac{4a^2b}{a^4+b^4+1}&\le\sqrt{2}\\[10pt] \text{as re}&\text{quired} \end{align*}