<s>[INST] 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)
 [/INST] Maybe a bit late answer but i thought it might be worth mentioning it.
Here is a direct elementary calculation of the maximum.
Let $x\geq 0$:
We only have to find the maximum of
$$\frac{x^3+x}{(x^2+x+1)^2} = \frac{x(x^2+1)}{(x^2+x+1)^2}$$
Simple rearrangements give
\begin{eqnarray*} \frac{x(x^2+1)}{(x^2+x+1)^2}
& = & \frac{x(x^2+x+1) -x^2}{(x^2+x+1)^2} \\
& = & \frac{x}{x^2+x+1} -\left(\frac{x}{x^2+x+1}\right)^2 \\
\end{eqnarray*}
Now, we have to maximize $t-t^2$ with $$0\leq t= \frac{x}{x^2+x+1} \stackrel{AM-GM}{\leq} \frac 13$$
with equality on the RHS for $x=1$.
The maximum of $t-t^2$ (a downward parabola with vertex at $t=\frac 12$) on $[0,\frac 13]$ is attained for $t=\frac 13$.
Hence,
$$\max_{x\geq 0}\frac{x^3+x}{(x^2+x+1)^2} = \frac 13 - \frac 19 = \frac 29 \Rightarrow \max_{x\geq 0}f(x) = \frac{\sqrt 2}{3}$$
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