Log sum inequality

Let ${\displaystyle a_{1},\ldots ,a_{n}}$ and ${\displaystyle b_{1},\ldots ,b_{n}}$ be nonnegative numbers. Denote the sum of all ${\displaystyle a_{i}}$s by ${\displaystyle a}$ and the sum of all ${\displaystyle b_{i}}$s by ${\displaystyle b}$. The log sum inequality states that

${\displaystyle \sum _{i=1}^{n}a_{i}\log {\frac {a_{i}}{b_{i}}}\geq a\log {\frac {a}{b}},}$

with equality if and only if ${\displaystyle {\frac {a_{i}}{b_{i}}}}$ are equal for all ${\displaystyle i}$, in other words ${\displaystyle a_{i}=cb_{i}}$ for all ${\displaystyle i}$.^[1]

(Take ${\displaystyle a_{i}\log {\frac {a_{i}}{b_{i}}}}$ to be ${\displaystyle 0}$ if ${\displaystyle a_{i}=0}$ and ${\displaystyle \infty }$ if ${\displaystyle a_{i}>0,b_{i}=0}$. These are the limiting values obtained as the relevant number tends to ${\displaystyle 0}$.)^[1]

Notice that after setting ${\displaystyle f(x)=x\log x}$ we have

${\displaystyle {\begin{aligned}\sum _{i=1}^{n}a_{i}\log {\frac {a_{i}}{b_{i}}}&{}=\sum _{i=1}^{n}b_{i}f\left({\frac {a_{i}}{b_{i}}}\right)=b\sum _{i=1}^{n}{\frac {b_{i}}{b}}f\left({\frac {a_{i}}{b_{i}}}\right)\\&{}\geq bf\left(\sum _{i=1}^{n}{\frac {b_{i}}{b}}{\frac {a_{i}}{b_{i}}}\right)=bf\left({\frac {1}{b}}\sum _{i=1}^{n}a_{i}\right)=bf\left({\frac {a}{b}}\right)\\&{}=a\log {\frac {a}{b}},\end{aligned}}}$

where the inequality follows from Jensen's inequality since ${\displaystyle {\frac {b_{i}}{b}}\geq 0}$, ${\displaystyle \sum _{i=1}^{n}{\frac {b_{i}}{b}}=1}$, and ${\displaystyle f}$ is convex.^[1]

The inequality remains valid for ${\displaystyle n=\infty }$ provided that ${\displaystyle a<\infty }$ and ${\displaystyle b<\infty }$.^{[citation needed]} The proof above holds for any function ${\displaystyle g}$ such that ${\displaystyle f(x)=xg(x)}$ is convex, such as all continuous non-decreasing functions. Generalizations to non-decreasing functions other than the logarithm is given in Csiszár, 2004.

Another generalization is due to Dannan, Neff and Thiel, who showed that if ${\displaystyle a_{1},a_{2}\cdots a_{n}}$ and ${\displaystyle b_{1},b_{2}\cdots b_{n}}$ are positive real numbers with ${\displaystyle a_{1}+a_{2}\cdots +a_{n}=a}$ and ${\displaystyle b_{1}+b_{2}\cdots +b_{n}=b}$, and ${\displaystyle k\geq 0}$, then ${\displaystyle \sum _{i=1}^{n}a_{i}\log \left({\frac {a_{i}}{b_{i}}}+k\right)\geq a\log \left({\frac {a}{b}}+k\right)}$. ^[2]