Finding max{f(x)} without derivative Consider the function $f(x)=\frac{\sqrt{x^3+x}}{x^2+x+1}$
the question is about: to find max{f} without using derivative.
I can find max with derivation and it is not hard to find. it is $f'(x)=0 \to x=1 $ so $max\{f\}=\frac{\sqrt 2}{3}$
but I am looking for an idea to do as the question said.
I am thankful for any help.(because I got stuck on this problem)

Using AM-GM, we have
\begin{align*}
 \frac{\sqrt{x^3 + x}}{x^2 + x + 1} &= \frac{\sqrt{2}\sqrt x \sqrt{\frac{x^2 + 1}{2}}}{x^2 + x + 1}\\
 &\le \frac{\sqrt{2}\cdot\frac12\left(x + \frac{x^2 + 1}{2}\right)}{x^2 + x + 1}\\
 &= \frac{\sqrt 2 (x^2 + 2x + 1)}{4(x^2 + x + 1)}\\
 &= \frac{\sqrt{2}}{3} - \frac{\sqrt 2 (x - 1)^2}{12(x^2 + x + 1)}\\
 &\le \frac{\sqrt{2}}{3}.
\end{align*}
Also, when $x = 1$,
we have $\frac{\sqrt{x^3 + x}}{x^2 + x + 1} = \frac{\sqrt 2}{3}$.
Thus, the maximum of $\frac{\sqrt{x^3 + x}}{x^2 + x + 1}$ is $\frac{\sqrt 2}{3}$.
