<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] Here is a proof that the maximal $x$ must be $\le 9$.
We consider $a = 1 + u$, $b = 1$ and compute the Taylor development of
$$
 f(u) = \frac 12 \left( (u+1)^2 + \frac{1}{u+1}\right) -
 \left( \frac{(1+u)^x + 1}{2}\right)^\frac 1x
$$
at $u = 0$.
From the geometric series
we get
$$
 \frac 12 \left( (u+1)^2 + \frac{1}{u+1}\right) =
 1 + \frac 12 u + u^2 + O(u^3)
$$
and from the binomial series we get
$$
(1+u)^x + 1 = 2 + xu + \frac{x(x-1)}{2} u^2 + O(u^3)
$$
and then
$$
\left( \frac{(1+u)^x + 1}{2}\right)^\frac 1x = 1 + \frac 12 u + \frac{x-1}{8}u^2 + O(u^3)
$$
It follows that
$$
 f(u) = \frac{9-x}{8} u^2 + O(u^3) \text{ for } u \to 0 \, .
$$
If $x > 9$ then $f(0) =f'(0) = 0$ and $f''(0) < 0$, i.e.
$f$ has a strict local maximum at $u = 0$, so that the desired
inequality does not hold for $a = 1+u$, $b=1$ with $u \ne 0$ 
sufficiently small.
</s>