Suppose we have $x_1\ln x_1+x_1^2=x_2\ln x_2 +x_2^2$ with $0<x_1<x_2$. Then
$x_1\ln x_1+x_2\ln x_2 +x_1+x_2+2x_1^2+2x_2^2>0$?
I am not sure the conclusion is true or wrong. From math software, it seems it is true. I have no idea how to deal with the problem, I am willing to prove the statement is true, but no result.
Any comment would be helpful.
Here's a proof with one step (in italics) missing; I think that step should be easier than the original problem.
Let $f(x) = x\ln x+x^2$ and $g(x) = x\ln x+x+2x^2$; also define $c\approx 0.2315$ to be the root of $f'(x) = 2x+\ln x+1$. Find a way to prove that $f(2c-x) < f(x)$ for $0<c<x$.
Choose $0<x_1<x_2$ so that $f(x_1)=f(x_2)$; then in fact $0 < x_1 < c < x_2$. Since $f(2c-x_1) > f(x_1)$, it follows that $x_2 > 2c-x_1$. Thus, since $g(x)$ is increasing for $x>c$, to prove that $g(x_1)+g(x_2)>0$ it suffices to prove that $g(x_1) + g(2c-x_1) > 0$ for $0 < x_1 < c$. But this follows from the fact that the derivative of the function $g(x) + g(2c-x)$ turns out to be $8(x-c) + \ln(x/(2c-x))$, which is negative for $0<x<c$, while $g(c) + g(2c-c)=0$.