# Callebaut's Inequality

Callebaut's Inequality states that for $1\ge x\ge y\ge 0,$ $$\sum_{i=1}^{n}a_{i}^{1+x}b_{i}^{1-x}\sum_{i=1}^{n}a_{i}^{1-x}b_{i}^{1+x}\ge\sum_{i=1}^{n}a_{i}^{1+y}b_{i}^{1-y}\sum_{i=1}^{n}a_{i}^{1-y}b_{i}^{1+y}.$$

It can be considered as an interpolation or a refinement of Cauchy-Schwarz, which is the $x=1,y=0$ case.

## Proof

Let $f(t):= \sum_{i=1}^{n}a_{i}^tb_{i}^{2-t}$. Then by Hölder, $f(1+y) \cdot f(1-y) \ge f(1)^2$, further (because $\frac yx\le 1$) $f(1-x)^{\frac yx}\cdot f(1)^{1-\frac yx} \ge f\left(\frac yx(1-x)+1-\frac yx\right)=f(1-y)$ and $f(1+x)^{\frac yx}\cdot f(1)^{1-\frac yx} \ge f\left(\frac yx(1+x)+1-\frac yx\right)=f(1+y)$.

Raising these three respectively to the $(\frac xy-1)$th, $\frac xy$th, $\frac xy$th power, we get $f(1+y)^{\frac xy-1}\cdot f(1-y)^{\frac xy-1} \ge f(1)^{2(\frac xy-1)}$ $f(1-x)\cdot f(1)^{\frac xy-1} \ge f(1-y)^{\frac xy}$ $f(1+x)\cdot f(1)^{\frac xy-1} \ge f(1+y)^{\frac xy}$

Multiplying the last three lines yields $f(1+x)f(1-x)\ge f(1+y)f(1-y)$ as required.