# 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.

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.

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. $$\qquad \Box$$