I came across the following inequality and I would like to know if it is true
$$(x^xy^{-y})^\frac{1}{y-x}(x-y)^2\leq x+y,$$
which is the same as proving
$$e^{-\frac{x\log x-y\log y}{x-y}}(x-y)^2\leq x+y.$$
When I do the simulations it seems it really holds, for any $x,y\geq 0\ $, ($x\ne y$).
I tried to use geometric-arithmetic mean inequality, i.e.,
$$(x^xy^{-y})^\frac{1}{y-x}=\left(\left(\frac{1}{x}\right)^{-x}\left(\frac{1}{y}\right)^{y}\right)^\frac{1}{y-x}\leq \frac{-x\frac{1}{x}+y\frac{1}{y}}{y-x}=0\leq x+y,$$
but obviously this is a wrong approach ("weights" $-x$, $-y$, can not be negative).
Do you have any idea how to prove it? Thanks.
Answer
Note that your formula show that your inequality is equivalent to
$$2\log |x-y|\leq \log (x+y)+\frac{x\log x-y\log y}{x-y}$$
Wlog, we may suppose that $x>y>0$, put $x=y+u$.
We have
$$x\log x=(y+u)\log(y+u)=y\log y+y\log (1+u/y)+u\log u+u\log(1+y/u)$$
and
$$\log (x+y)=\log (2y+u)=\log u+\log (1+2y/u)$$
Hence we have to show:
$$2\log u\leq \log u+\log (1+2y/u)+(y/u)\log (1+u/y)+\log u+\log(1+y/u)$$
or
$$0\leq \log (1+2y/u)+(y/u)\log (1+u/y)+\log(1+y/u)$$
and this is clear.
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