Use Yang's inequality to prove: if $a,b\geq 1$, then $ab\leq e^{a-1}+b\ln b$.
Here is the definition of Yang's inequality the exercise gives:
Let $f\in C^1[0,+\infty)$ be strictly monotone increasing, $f(0)=0$, $a,b>0$, then we have $$ ab\leq\int_0^a f(x)\mathrm{d}x+\int_0^b g(y)\mathrm{d}y $$ where $g(y)$ is the inverse function of $f(x)$.
I tried it by setting $f(x)=e^{x-1}-e^{-1}$(to satisfy $f(0)=0$ and the term $e^{a-1}$). Then I get $$g(y)=\ln(e^{-1}+y) +1$$ $$\int_0^x f(x)=e^{x-1}-e^{-1}x-e^{-1}$$ and $$\int_0^y g(y)=(e^{-1}+x)\ln(e^{-1}+x)+e^{-1}.$$ Substitute the Yang's inequality I get
$$ ab\leq e^{a-1}-e^{-1}a+(e^{-1}+b)\ln(e^{-1}+b). $$ This cannot solve my question. Is there anything wrong in my derivation or I just choose the wrong function $f(x)$?
I even tried the function $e^x-1$ for $a$ and $eb$ still not work.
One Source
Functions $$ f(x) = e^x - 1, $$ with antiderivative $$ e^x - x - 1 $$ that also vanishes at the origin, to fit the given integral. Then $$ g(y) = \log (1+y), $$ antiderivative $$ (1 + y)\log (1 + y) - y, $$ vanishes at zero.
Apply inequality for
$$ a-1, b-1 $$ which are positive, as required.