elementary inequality involving exp and ln

Question

Suppose $x$ and $y$ are real numbers satisfying $x\geq 0$ and $y\geq 1$. Is the following inequality true?

$xy \leq e^x + y \ln (y)$

If so, is there a reference or proof?

Answer 1

Writing your inequality as $(x-\ln y) y\le e^x$ and letting $z:=x-\ln y$, we want to prove that $$ zy \le e^{z+\ln y} = e^zy, $$ which is obviously true.

Answer 2

Let $f(x)=e^x+y\ln{y}-xy.$

Thus, $f'(x)=e^x-y,$ which says $$f(x)\geq f(\ln{y})=y+y\ln{y}-y\ln{y}=y>0.$$