For which $(n,m)$ holds $x^ny^m + x^my^n \le 2$, where $x+y = 2$? For which $(n,m)$ holds $x^ny^m + x^my^n \le 2$, where $x+y = 2$ and $x,y \ge 0$? 
It is  $n,m \ge 0$ where $n$ and $m$ do not have to be integers.
The conjectured answer is: for $|n-m | \le \sqrt{n+m}$.
The questions looks for a formal proof.                   
Also a proof for integer  $n$ and $m$ would be acknowledged.

In the following  is an informal approach. Of course, a shorter / more concise  way
towards a solution is appreciated.
Let's start with some limiting cases for illustration.
For $m=n$, the conjecture is always true.
This is correct, since then we need to prove
$x^ny^n  \le 1$ or $\sqrt{xy}  \le 1$ which is always true by AM-GM-inequality, since
$\sqrt{xy}  \le \frac{x+y}{2} = 1$.
As another extreme case, for $n \to 0$ and $m > 0$  the conjecture gets $m\le \sqrt{m}$ or $m \le 1$.  This is also correct, since  the inequality  gets $y^m + x^m \le 2$. Let  $x = 1+d$ and $y = 1-d$. Then we have $f(d) = (1+d)^m + (1-d)^m $ which has one extremum  at $d=0$. The second derivative is $f´´(d=0) = 2 (m - 1) m$ so indeed for $m < 1$ this is a maximum and $y^m + x^m \le 2$.
Now for the general case. Homogenize the inequality by
$(x^ny^m + x^my^n) 2^{m+n} \le 2 (x+y)^{m+n}$ or
$$
f = 2 (1+a)^{m+n} - 2^{m+n} (a^{b(m+n)} + a^{(1-b)(m+n)}) \ge 0
$$
where $a = y/x$ and $b = \frac{m}{m+n}$. The case  $b = 0.5$ corresponds to $m=n$ as discussed  above.
Moving away from $b=0.5$, a solution $f = 0$ for a range of $m+n$ exists for  specific $a$. E.g. for $b= 0.4$ we have
$$
f = 2 (1+a)^{m+n} - 2^{m+n} (a^{0.4(m+n)} + a^{0.6(m+n)}) \
$$
and solutions $f = 0$ are for example (using Matlabs fzero)
$(a,m+n) = (0.45, 317.8779) ; (0.5, 66.4909) ; (0.55, 43.6458) ; (0.6, 35.4395) ; (0.65, 31.3039) ; (0.75,  27.3412) ; (0.85,  25.6856) ; (0.95,  25.0660); (0.99,  25.0025) $
For  $m+n > 25$, there is always  a solution $f = 0$ so the inequality cannot hold.
We see that, generally, for the inequality to hold, $m+n$ must be below a maximum  which is attained as $a \to 1$. To compute that maximum, let $a = 1 - d$ and expand $f$ for small $d$. One obtains to lowest order in $d$:
$
f = 2^{m+n - 2} (m+n)  (1 - (m+n) (1 - 2 b)^2 ) \; d^2
$
So $f \ge 0$ for $1 - (m+n) (1 - 2 b)^2  \ge 0$. This gives   $1 \ge (m+n) (\frac{m-n}{m+n})^2$ or $|n-m | \le \sqrt{n+m}$ as conjectured.
For the example case $b=0.4$ discussed above, the limit corresponds to $0.4 = \frac{m}{m+n}$ and $|n-m | = \sqrt{n+m}$, which is solved by $m = 10$ and $n = 15$, leading to the limit at $m+n = 25$ as observed numerically above.
 A: The asked function, taking into account the boundary condition, can be cast as
$f = -2 + (1+q)^n (1-q)^m + (1+q)^m(1-q)^n \le 0$ where $-1 < q < 1$. A series expansion reads
$ f \simeq [(m-n)^2 - (m+n)] q^2 + \cal{O}(q^4)$
So a necessary requirement for $f \le 0$ is $|m-n| \le \sqrt{m+n}$ which is exactly the conjecture.
What needs to be proved is the sufficiency of  this. Note that $f(q=0) = 0$   and $f(q=\pm 1) = -2$. Let us regard the condition  $|m-n| < \sqrt{m+n}$. For small $q$, $f$ will then fall  with $q$. Since $f(q=\pm 1) = -2$, the function can either continue falling with $q$ which is the desired case, or rise again which offers the opportunity that $f > 0$ will be reached which is the non-desired case. For the latter to happen, $f'(q) = 0$ must be observed. Let us show that this will not happen.
W.l.o.g. let $m \ge n$ and $k = m-n$, then equivalently show
$f = -2 + (1-q^2)^n\Big[ (1+q)^k + (1-q)^k \Big] \le 0$. From this formulation,
We have
$$
f'(q) = (1 - q^2)^{n-1}  \Big[-q (2 n + k) ((1 - q)^k + (q + 1)^k)  + k (-(1 - q)^k + (q + 1)^k) )\Big]
$$
For $f'(q) = 0$ we need
$$
g(q) = \frac{-(1 - q)^k + (q + 1)^k}{(1 - q)^k + (q + 1)^k}= q\frac{2n +k}{k}
$$
Now in the LHS formulation, $g(q)$ is a sigmoid function, point-symmetric to $q=0$,  which is concave for $q \ge 0$. Using the linearized behavior of $g(q)$ at $q=0$,  we have the condition
$$
q\frac{2n +k}{k} = g(q) \le k q 
$$
or, reinstating $k$, we have $ \frac{n+m }{m-n} \le m-n
$ which is  $|m-n| > \sqrt{m+n}$. However, this will not happen, since we are in the complementary realm $|m-n| < \sqrt{m+n}$. This proves that $f$ will indeed fall  with $q$ everywhere. $\qquad \Box$
