Show g' is increasing and deduce the inequality Let $g :=$ -ln $x$ where $x>0$. Show $g′$ is increasing on $(0,∞)$ and deduce that if $n ∈ \mathbb{N}$, $x_1,...,x_n$ are positive and $s_1, . . . , s_n$ are non-negatives with $\sum_{k=1}^n s_k = 1$, then
$x_1^{s_1}...x_n^{s_n}\leq s_1x_1+...s_nx_n$.
I understand that $g''=1/x^2$ and is positive on $(0,∞)$, so $g'$ is increasing. But I'm not sure how the rest follows.

Hint: $g(x)=-\ln x$ then $g'$ increasing since $g''(x)=1/x^2>0$ on $(0,\infty)$. Which is to say log is concave. Notice that $\ln(x_1^{s_1}...x_n^{s_n})=s_1\ln x_1+...+s_n\ln x_n\leq\ln(s_1x_1+...s_nx_n).$(by concavity) Then ...
