I have found this method in solving this kind of inequality :
Let $x,y>0$ such that $x+y=1$ then we have : $$x^{2y}+y^{2x}\leq 1$$
Well now we want to solve the general problem :
$$a+b\leq 1$$
Under some constraint on $1>a,b>0$
Well a bit of algebra and trigonometric's formulae shows that it's equivalent to :
$$\sin^2\Big(a\frac{\pi}{2}\Big)+\sin^2\Big(b\frac{\pi}{2}\Big)\leq 1$$
But :
$$\sin^2\Big(a\frac{\pi}{2}\Big)+\sin^2\Big(b\frac{\pi}{2}\Big)\leq \cos^2\Big(a\frac{\pi}{2}\Big)+\sin^2\Big(a\frac{\pi}{2}\Big)$$
Or : $$\sin^2\Big(b\frac{\pi}{2}\Big)\leq \cos^2\Big(a\frac{\pi}{2}\Big)\quad (1)$$
Or :
$$\Big(\sin\Big(b\frac{\pi}{2}\Big)-\cos\Big(a\frac{\pi}{2}\Big)\Big)\Big(\sin\Big(b\frac{\pi}{2}\Big)+\cos\Big(a\frac{\pi}{2}\Big)\Big)\leq 0$$
The problem is :
$$\Big(\sin\Big(b\frac{\pi}{2}\Big)-\cos\Big(a\frac{\pi}{2}\Big)\Big)$$
Or :
$$\Big(\cos\Big((b-1)\frac{\pi}{2}\Big)-\cos\Big(a\frac{\pi}{2}\Big)\Big)$$
Now we use the fact :
$$\cos(p)-\cos(q)=-2\sin\Big(\frac{p+q}{2}\Big)\sin\Big(\frac{p-q}{2}\Big)$$
We get :
$$-2\sin\Big(\frac{\pi(b+a-1)}{4}\Big)\sin\Big(\frac{\pi(b-1-a)}{2}\Big)$$
Now it's clear that it's equivalent to the original problem.
Now the idea is using power series: factorize the expression. We have :
$$\cos^2(x)=\sum_{n\geq 2} \frac{(2 i)^{-2 + n} (1 + (-1)^n) (-\frac{π}{2} + x)^n}{n!}$$ See here
And :
$$\sin^2(x) = - \sum_{k=1}^{\infty} \frac{(-1)^k 2^{-1 + 2 k} x^{2 k}}{(2 k)!}$$
See here
Well now making the difference in the inequality $(1)$ we can factorize to get someting like :
$$0\leq \frac{\pi}{2}(a-1-b)\Big(P\Big(\frac{\pi}{2}a,\frac{\pi}{2}b\Big)\Big)$$
Here I have used the fact :
$$x^n-y^n=(x-y)(Q(x,y))$$
With $n\geq 1$ a natural number .
Now the problem is the :
$$P(\frac{\pi}{2}a,\frac{\pi}{2}b)\quad (2)$$
My questions : Is it just unsuable ? Can we have more informations on $(2)$ ,I mean as it is equal to zero ? Can we improve the situation ? Can we have a good estimation of $(2)$ ?
Thanks in advance !!