<s>[INST] Show that $\frac{1}{b-a_{1}} + \frac{1}{b-a_2}+ ...+ \frac{1}{b-a_n} \geq \frac{n}{b-\frac{1}{n}(a_1+a_2+...+a_n)}$ Let, $b> \max\{a_1,a_2,...,a_n\}.$ Show that $\frac{1}{b-a_{1}} + \frac{1}{b-a_2}+ ...+ \frac{1}{b-a_n} \geq \frac{n}{b-\frac{1}{n}(a_1+a_2+...+a_n)}$
f is convex if $f(t_1x_1+t_2x_2+...+t_nx_n)\leq t_1f(x_1)+t_2f(x_2)+...+t_nf(x_n)$
Let, $t_i = 1/n \Rightarrow \sum\limits_{i=1}^n t_i = 1$
Then, $f(\frac{a_1+a_2+...+a_n}{n}) \leq \frac{1}{n}[f(a_1)+f(a_2)+...+f(a_n)]$
NTS: $\frac{1}{b-\frac{1}{n}(a_1+a_2+...+a_n)} \leq \frac{1}{n}[\frac{1}{b-a_1}+...+\frac{1}{b-a_n}]$
Let$f(x) = \frac{1}{b-x}$ In order to prove the inequality, we have to show that $f''(x) \geq 0.$ So, $f''(x) = \frac{2}{(b-x)^3}$
Now,  how can we show that $f''(x) \geq 0$? $b$ can be a negative number, can it? 
 [/INST] I would use the function $g(x)=\frac{1}{x}$ defined on $(0,\infty)$; you can readily verify that $g''(x)>0$ on its domain. (The issue with your $f$ is that $f''(x)>0$ only for $x<b$.) The rest of the argument proceeds similarly: convexity gives
$$
\frac{1}{n}\times\text{LHS}=\frac{1}{n}\sum_{i=1}^ng(b-a_i)\geq g\left(b-\sum_ia_i/n\right)=\frac{1}{n}\times\text{RHS}\implies\text{LHS}\geq \text{RHS}.
$$
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