Callebaut's Inequality

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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.

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