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$$\frac x{(x^2+1)^{\frac32}}$$

I want to find the maximum value of this expression, and the value of x at that time.
but not by derivating the function, instead using AM-GM inequality.

I found that the maximum value is $\frac{2}{3\sqrt{3}}$ and the value of x is $\frac1{\sqrt2}$ by wolfram alpha, but I want to prove it using AM-Gm inequality. Thanks in advance.

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    $\begingroup$ Try the antiderivative it's simple if you want to check your result ;-).Good day dissolve . $\endgroup$ Aug 10, 2020 at 10:59

1 Answer 1

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By AM-GM, $$\frac{x^2+1}{3}=\frac{x^2+\tfrac{1}{2}+\tfrac{1}{2}}{3}\ge\sqrt[3]{x^2/4}$$ so $$\frac{x}{(x^2+1)^{3/2}}\le\frac{2}{\sqrt{3^3}}$$ with equality when $x^2=1/2$, that is $x=1/\sqrt2$.

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