prove that inf $f\geq 0$ if and only if $ae^{1+b}\geq 1$.

Question

Ive got a function $f(x)=a+bx+xlnx$ for all $x\in(0,\infty)$ and I need to prove that inf $f\geq 0$ if and only if $ae^{1+b}\geq 1$.

I have already proved that $f$ is convex and have found a formula for inf, inf $f=$max$\{b\in \mathbb{R}: v\leq x$ for all $x\in f\}$ but I am unsure how to implement the question to show what is required. Any help will be great.

Answer 1

We want $(b+\ln x)x\geq -a$ for all $x >0$ which means $\inf \{(b+\ln x)x: x>0\} \geq -a$. By differentiation we see that the infimum is attained at $x=e^{-1-b}$. [The function is decreasing up to this point and increasing after that]. Thus the condition we need is $(b+(-1-b)) e^{-1-b} \geq -a$ which says $ae^{1+b } \geq 1$.