Is the inequality $\:\frac{r^2}s +\frac{s^2}r\geqslant 2\max\{r,s\}\:$ a valid one? I'm not sure if
$$\frac 12\left(\frac{r^2}s + \frac{s^2}r\right)\;\geqslant\;
\max\{r,s\}$$
holds true for all $\,r,s\in\mathbb R^{>0}$.
I think so $\,\ddot\smile\:$ but
Do you have a good proof for it?
$\ddot\frown\:$ The cyclic analogue in three variables
$$\max \big\{u,v,w\big\} \;\overset{\displaystyle ?}{\leqslant}\;
\frac 13\left(\frac{u^2}v + \frac{v^2}w + \frac{w^2}u\right)$$
certainly/already fails because evaluating the RHS at $\,(u,v,w)=(3,2,1)\,$ yields $\,2\tfrac{17}{18}$.
 A: Note that if $\,1<\frac rs<1.618$, then
$$\frac 12\left(\frac{r^2}s + \frac{s^2}r\right)\;\geqslant\;
\max\{r,s\}$$ is not true.
For example put $\,r=1.5 s\,$ so we have 
$$\frac 12\left(\frac{(1.5s)^2}s + \frac{s^2}{1.5s}\right)\;\geqslant\;\max\{1.5s,s\}\\
\frac 12\left(2.25s +.666...s \right)\;\geqslant\;\max\{1.5s,s\}\\
(2.25s +.666...s )\;\geqslant\;2\max\{1.5s,s\}\\[2ex]
2.91666s\;\color{red}\ngeqslant\;3s\\$$
A: 
Your Claim is False See below: Indeed, 
  \begin{split}
&&\frac 12\left(\frac{r^2}s + \frac{s^2}r\right)\;\geqslant\;
\max\{r,s\}\\ 
&\Longleftrightarrow& \frac 12\left(\frac{r^2}s + \frac{s^2}{r}\right) \;\geqslant\; r \quad and \quad\frac 12\left(\frac{r^2}{s} + \frac{s^2}r\right)\;\geqslant\;s \\  &\Longleftrightarrow& \left(\frac{r^2}s + \frac{s^2}{r}\right) \;\geqslant\; 2r \quad and \quad\left(\frac{r^2}{s} + \frac{s^2}r\right)\;\geqslant\;2s \\&  \Longleftrightarrow& \left(\frac{r}s + \frac{s^2}{r^2}\right) -2\;\geqslant\;
0 \quad and \quad\left(\frac{r^2}{s^2} + \frac{s}r\right)-2\;\geqslant\;
0\\& \Longleftrightarrow &f(t)\ge 0 \quad and \quad f(\frac{1}{t})\ge 0
\end{split}

Where $t=\frac{s}{r}$ and  $f(t) = t^2 +\frac{1}{t} - 2$ , for $t>0$. Let study $f(t)$ for $t>0$. 
We have
$$f'(t) = 2t -\frac{1}{t^2} = 0\Longleftrightarrow t= \frac{1}{\sqrt[3]{2}}$$
where $$\lim_{t\to 0}f(t)=\lim_{t\to \infty}f(t)=\infty$$
Therefore $f(\frac{1}{\sqrt[3]{2}})$ is the only minimum of $f$ on $(0,\infty)$ that is,
 $$\forall t>0,~~  f(t) \ge f(\frac{1}{\sqrt[3]{2}}) = 3*2^{-\frac{2}{3}} -2$$
But $$3*2^{-\frac{2}{3}} -2<0 $$ 
therefore there exists $0<a<1$ such that $f(a) = f(1) = 0$  and $f(t)< 0$ for $t\in(a,1)$ in particular 
$$f(\frac{1}{\sqrt[3]{2}}) = 3*2^{-\frac{2}{3}} -2<0.$$ 
Moreover, we have, $$a=\frac{1+\sqrt{5}}{2}\qquad \text{that is,}\qquad f( \frac{-1+\sqrt{5}}{2}) = f(1)=0 $$
and $$f(t)< 0 \qquad \text{for}\qquad \frac{-1+\sqrt{5}}{2}<t<1.$$
Hence your claim is FALSE
