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)
 A: 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 $\frac{\sqrt 2}{3}$.
A: 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,
$$\max_{x\geq 0}\frac{x^3+x}{(x^2+x+1)^2} = \frac 13 - \frac 19 = \frac 29 \Rightarrow \max_{x\geq 0}f(x) = \frac{\sqrt 2}{3}$$
A: Just to add, if you already have a guess, then proving it is quite easy.  i.e. given $x$ and hence $f(x)$ is non-negative, we can square things and factor the resulting polynomial (as one root is actually already known to us ;)
$$\frac{\sqrt{x^3+x}}{x^2+x+1} \leqslant \frac{\sqrt2}3 \iff (x-1)^2(2x^2-x+2)\geqslant 0$$
The latter is obvious as the quadratic factor is always positive.  Now all that remains is to show there is equality when $x=1$.

Just for completeness, providing another way with just AM-GM:  with $t^2=x+\frac1x$, we can write the expression as
$$f = \frac{\sqrt{x^3+x}}{x^2+x+1} =\frac{x\sqrt{x+\frac1x}}{x(x+\frac1x+1)} = \frac{t}{t^2+1} = \frac1{t+\frac1t}$$
Now, $t^2=x+\frac1x \geqslant 2$ by AM-GM, with equality iff $x=1$.  Hence to maximise $f$, we need to find the minimum of $t+\frac1t$, when $t\geqslant \sqrt2$. However, in this domain, we can write using AM-GM and the obviously increasing function $-\frac1t$,
$$t+\frac1t=t + \frac{2}t-\frac1t \geqslant 2\sqrt2-\frac1t \geqslant 2\sqrt2-\frac1{\sqrt2} =  \frac3{\sqrt2}$$
As equality here is when $t=\sqrt2 \iff x=1$, we have a consistent point of equality.
Thus $f = \dfrac1{t+\frac1t} \leqslant \dfrac{\sqrt2}3$
A: By AM-GM inequality we have that
$$\frac{\sqrt{x^3+x}}{x^2+x+1} \leq \frac{\sqrt{x^3+x}}{3x} = \frac{\sqrt{2}}{3}$$
since the AM-GM equality is attained when $x^2=x=1$
