Callebaut's Inequality
Callebaut's Inequality states that for
It can be considered as an interpolation or a refinement of Cauchy-Schwarz, which is the case.
Proof
Let . Then by Hölder,
, further (because
)
and
.
Raising these three respectively to the th,
th,
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 as required.