It's a following problem:
Let $a,b>0$ such that $a+b=1$ then we have : $$(a^b+b^a)(a^{b+1}+b^{a+1})\geq 1$$
I try to use Tchebytchev inequality to get a sum but it's too weak as approximation .
I try otherwise something like : $$(a^b+b^a)\geq \sqrt{a}+\sqrt{b}$$
But it doesn't work.
I don't know if it's relevant but the function :
$$f(x)=(x^{(1-x)+1}+(1-x)^{x+1})$$ is convex on $[0,1]$
Maybe we need a really new approach to solve it .
Thanks a lot for your time and patience .
Probably not finished.
Because of the symmetry, we only need to focus on the range $0\leq a \leq \frac 12$ after having replaced $b$ by $(1-a)$.
Now, consider the function $$f(a)=(a^b+b^a)(a^{b+1}+b^{a+1})\qquad \text{where} \qquad b=1-a$$ The function value is $1$ for $a=0$ and $a=\frac 12$.
Expansions at both ends of the interval are $$f(a)=1-(\log (a)+2)a^2 +\left(\frac{\log ^2(a)}{2}+1\right)a^3 +O\left(a^4\right)$$ $$f(a)=1+ \left(\log ^2(2)+4\log (2)-3\right)\left(a-\frac{1}{2}\right)^2+O\left(\left(a-\frac{1}{2}\right)^4\right)$$ At least we know that the function starts and ends with valuee equal to $1$.
Performing a similar expansion around $a=\frac{1}{4}$ up to $O\left(\left(a-\frac{1}{4}\right)^4\right)$ (too messy formula to be reported here shows that the function is maximum at $a=0.194853$ (the exact solution being $a=0.193776$). For this approximate value of $a$, the maximum value is predicted to be $1.01083$ which is exactly the value at the exact maximum.