Puzzled by [this still open question](https://mathoverflow.net/questions/193753), I tried comparing the arithmetic mean $A(x,y)=(x+y)/2$ with a mean intermediate between a geometric-type mean $G(X)=(x^a y^{1-a}+x^{1-a} y^a)/2\;$ for $0\le a \le 1$, and an $L^p$-type mean $L(x,y)=(x^p+y^p)^{1/p}\;$ for $p\ge 1$.


Notice that $G(x,y)\le A(x,y)\le L(x,y)\;$ for $x\ge0, y\ge0$. Then define the mixed mean


$$M\_{a,p}(x,y)=\Big(\frac{(x^a y^{1-a})^p+(x^{1-a} y^a)^p}{2}\Big)^{\frac{1}{p}}$$


The following are easily proved:


* $M\_{a,p}\le M\_{b,q}\;$ if $\;1\le p\le q\;$ and $\;1/2\le a\le b\;$;
* given $1/2\le a< 1$ and $p\ge1$, then $\;A(x,y)>M\_{a,p}(x,y)\;$ for $x/y$ or $y/x$ large enough;
* the Hessian of $\;M\_{a,p}\;$ at $\;(x,y)=(1,1)\;$ is $\;\displaystyle\frac{p(p(2a-1)^2-1)}{8}\left( \begin{smallmatrix} 1&-1\\ -1&1\end{smallmatrix}\right)\;$, which is negative semidefinite for $p\le (2a-1)^{-2}\;$;


therefore


$$M\_{a,p}\le A=M\_{1,1}\;\; \text{on}\;\; \mathbb{R}\_{+}^2\implies p\le \frac{1}{(2a-1)^2}$$


**Question**: is the converse true? In other words, is $M\_{a,\frac{1}{(2a-1)^2}}\le A\;$?


After checking few numerical examples, I would guess a positive answer.


Notice that the more general comparison between $M\_{a,p}$ and $M\_{b,q}$ can be reduced easily to the comparison between some other $M\_{c,r}$ and $M\_{1,1}$.


In 4 variables, the local condition implied by the Hessian is not sufficient
(see example [here](https://mathoverflow.net/questions/193753/this-inequality-why-cant-solve-it-by-now-only-four-variables-inequality#comment483107_193753)).



Is this parametric inequality true?
Let me change the notation a bit by calling $m=\frac{1}{2a-1}>1$. Then your inequality reads
$$\left(\frac{x+y}{2}\right)^{m^2}\geq (xy)^{\binom{m}{2}}\cdot\left(\frac{x^m+y^m}{2}\right).$$
This is true for all $x,y\geq 0$, and I would love to see a slick elementary proof. However in the meantime notice that you can prove it by using Lagrange multipliers. Let's normalize to $x+y=1$ and check that the critical points of $(xy)^{\binom{m}{2}}\left(x^m+y^m\right)$ happen where $$(m-1)(x^{m+1}-y^{m+1})=(m+1)xy(x^{m-1}-y^{m-1}),$$
but it's not hard to check that this can only happen if $x=y$. And so by a simple check on the boundary values, one can conclude that your inequality is true.


