<s>[INST] Seeking the maximal parameter value s.t. two-variable inequality still holds Consider the expression
$$\frac 12\left(\frac{a^2}b+\frac{b^2}a\right)$$
in two variables $\,a,b\,$ residing in $\,\mathbb R^{>0}$.
The arithmetic mean $\,\frac{a+b}2\,$ is a lower bound for it
$\big[$ one has $\,a^2b\le(2a^3+b^3)/3\,$ by AM-GM, do the same with $\,ab^2$, then add and divide by $2ab\,$$\big]$,
but $\,\max\{a,b\}\,$ is not
$\big[$ choose $\,(a,b)=(3,2)\,$ for instance, the expression then evaluates to $\,2\tfrac{11}{12}\,\big]$.
Both arithmetic mean ($x=1$) and the maximum ($x=\infty$) are instances of the Hölder mean
$$\left(\frac{a^x+b^x}2\right)^{\frac 1x}\quad\text{with }\; x\,\in\,\{-\infty\}\cup\mathbb R\cup\{\infty\}$$
aka Power mean, known to be strictly increasing with $\,x\,$ if $\,a\ne b$, and my question is:

What is the maximal value of $\,x\,$ such that
  $$\left(\frac{a^x+b^x}2\right)^{\frac 1x}\;\leq\;\frac 12\left(\frac{a^2}b+\frac{b^2}a\right)$$
  holds for all $\,a,b>0\,$?

Let's pick up the specific "max counter-example" $\,(a,b)=(3,2)\,$.
The following plot screen-shot displays the zero, by courtesy of WolframAlpha:


Returning to the general case we can at least state that $\,x_{max}\geqslant 5\,$:
After taking the fifth power of the corresponding expression and clearing denominators (to arrive at the LHS as of below) I could find a
Certificate of positivity
$$\begin{eqnarray}
\left(a^3 + b^3\right)^5 -16a^5b^5\left(a^5 +b^5\right)\; & =\;\left(a+b\right)\left(a-b\right)^2\big[a^{12} + a^{11}b +2a^{10}b^2 +4a^9b^3 +8a^8b^4 \\
&  +8a^6b^6 +8a^4b^8 +4a^3b^9 + 2a^2b^{10} +ab^{11} +b^{12}\big] \\
& + 3a^3b^3\left(a^2+b^2\right)\left(a+b\right)^3\left(a-b\right)^4
\end{eqnarray}$$
It shows also that the equality case holds iff $\,a=b$.
 [/INST] The following inequality is true.
Let $a$ and $b$ be positive numbers.
Prove that
$$\frac{1}{2}\left(\frac{a^2}{b}+\frac{b^2}{a}\right)\geq\sqrt[9]{\frac{a^9+b^9}{2}}.$$
Indeed, we need to prove that
$$(a^3+b^3)^9\geq256(a^9+b^9)a^9b^9$$ or
$$(a^3+b^3)^8\geq256(a^6-a^3b^3+b^6)a^9b^9.$$
Now, let $a^6+b^6=2ua^3b^3.$  
Hence, $u\geq1$ and we need to prove that
$$(2u+2)^4\geq256(2u-1),$$ which is AM-GM:
$$(2u+2)^4=(2u-1+3)^4\geq\left(4\sqrt[4]{(2u-1)\cdot1^3}\right)^4=256(2u-1).$$
Done!
</s>