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.

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Callebaut's Inequality

Proof