Let $0<x<y$. Use the Mean Value Theorem to prove:
- $y\ln y - x\ln x\leqslant(y-x)(1+\ln y)$
- $e^{y^2}-e^{x^2}\leqslant (y-x)(x+y)e^{y^2}$
The first inequality is easy. Just consider the function $f\colon\mathbb{R}_{>0}\to\mathbb{R}, f(x)=x\ln x$, then by the MVT, for some $z\in (x,y)$, $$ f'(z)=1+\ln z=\frac{f(y)-f(x)}{y-x}=\frac{y\ln y-x\ln x}{y-x}. $$ Since $f'(x)=1+\ln x$ is monotonically increasing and $z<y$, it follows that $$ f'(y)=1+\ln y\geqslant f'(z) $$ which yields the inequality which was to be proven.
For the second inequality, $f(x)=e^{x^2}$ and thus $f'(x)=2xe^{x^2}$ so that with the same argument as above, I get $$ 2ye^{y^2}\geqslant\frac{e^{y^2}-e^{x^2}}{y-x} $$
I don't see where the factor $x+y$ comes into play. It seems to me that I need another estimate which replaces the factor $2y$ on the left-hand side by $x+y$, but since $2y\geqslant x+y$, I do not see a helpful upper estimate.
What am I overlooking?
Apply the mean-value theorem to $f(x) = e^x$ (instead of $e^{x^2}$): $$ e^{y^2}-e^{x^2} = f(y^2)-f(x^2) = (y^2-x^2) f'(z) = (y-x)(x+y)e^z $$ for some $z \in (x^2, y^2)$, and the desired inequality follows.