Puzzled by this still open question, 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).