Jensen's inequality
Prove that,
$$f(\sum^M_{m=1}\lambda_mx_m)\geq\sum^M_{m=1}\lambda_mf(x_m)$$
\(f()\) is any concave function and \(\sum^M_{m=1}\lambda_m=1\) for \(\{\lambda_m\}^M_{m=1} \geq 0\).
Similarly, the inequality gets reversed for convex function.
Proof
Given that, if \(a\leq c\leq b\), \(f(c)=f((1-\lambda)a+\lambda b)\geq (1-\lambda)f(a)+\lambda f(b)\). Recursively,
\begin{equation}
f(\sum^M_{m=1}\lambda_mx_m)=f(\lambda_1x_1+\sum^M_{m=2}\lambda_mx_m)=f\bigg(\frac{\lambda_1x_1}{\sum^M_{m=1}\lambda_m}+\frac{\sum^M_{m=2}\lambda_mx_m}{\sum^M_{m=2}\lambda_m}\frac{\sum^M_{m=2}\lambda_m}{\sum^M_{m=1}\lambda_m}\bigg)\\
= f\bigg(p_1x_1+(1-p_1)\frac{\sum^M_{m=2}\lambda_mx_m}{\sum^M_{m=2}\lambda_m}\bigg)\geq p_1f(x_1)+(1-p_1)f\bigg(\frac{\sum^M_{m=2}\lambda_mx_m}{\sum^M_{m=2}\lambda_m}\bigg) \\
= \lambda_1f(x_1)+\sum^M_{m=2}\lambda_mf\bigg(\frac{\sum^M_{m=2}\lambda_mx_m}{\sum^M_{m=2}\lambda_m}\bigg) = \lambda_1f(x_1)+\sum^M_{m=2}\lambda_mf\bigg(\frac{\lambda_2x_2+\sum^M_{m=3}\lambda_mx_m}{\sum^M_{m=2}\lambda_m}\bigg) \\
= \lambda_1f(x_1)+\sum^M_{m=2}\lambda_mf\bigg(\frac{\lambda_2x_2}{\sum^M_{m=2}\lambda_m}+\frac{\sum^M_{m=3}\lambda_mx_m}{\sum^M_{m=3}\lambda_m}\frac{\sum^M_{m=3}\lambda_m}{\sum^M_{m=2}\lambda_m}\bigg) \\
= \lambda_1f(x_1)+\sum^M_{m=2}\lambda_mf\bigg(p_2x_2+(1-p_2)\frac{\sum^M_{m=3}\lambda_mx_m}{\sum^M_{m=3}\lambda_m}\bigg) \\
\geq \lambda_1f(x_1)+\sum^M_{m=2}\lambda_m\bigg(p_2f(x_2)+(1-p_2)f\bigg(\frac{\sum^M_{m=3}\lambda_mx_m}{\sum^M_{m=3}\lambda_m}\bigg)\bigg) \\
= \lambda_1f(x_1)+\lambda_2f(x_2)+\sum^M_{m=3}\lambda_mf\bigg(\frac{\sum^M_{m=3}\lambda_mx_m}{\sum^M_{m=3}\lambda_m}\bigg) = \cdots \geq \sum^M_{m=1}\lambda_mf(x_m)
\end{equation}
References