I have to solve the following exercise.
If $a_{k}, b_{k} \geq 0, p>1, p=q(p-1)$,and define $F_{n}$ $$F_{n}=\left(\sum_{k=1}^{n} a_{k}^{p}\right)^{1 / p}\left(\sum_{k=1}^{n} b_{k}^{q}\right)^{1 / q}-\sum_{k=1}^{n} a_{k} b_{k}$$ research the $F_{n}$ monotony.
My attempt:
I found out the formula is similar to holder-inequality.But don't know what to do next.
Let $A=\sum_{k=1}^{n} a_{k}^{p}$, $B=\sum_{k=1}^{n} b_{k}^{q}$.
$F_{n+1}-F_n = \left(A+a^p_{n+1}\right)^{\frac{1}{p}}\left(B+b^q_{n+1}\right)^{\frac{1}{q}}-A^{\frac{1}{p}}B^{\frac{1}{q}}-a_{n+1}b_{n+1}\geq 0$
The last one is true because of Holder for $(A^{\frac{1}{p}},a_{n+1}), (B^{\frac{1}{q}},b_{n+1})$