How to prove that $$\sum\limits_{k=1}^n (\frac{a_1+a_2+\dots+a_k}{k})-\sum\limits_{k=1}^n a_k^2\leq \frac{n}{2}-\sum\limits_{k=1}^n\frac{1}{4k}?$$
In my book it is said that this comes from Cauchy-Schwartz inequality and from that $$\sum\limits_{k=1}^n (\frac{a_1+a_2+\dots+a_k}{k})\leq (2n-\sum_{k=1}^n\frac{1}{k})^\frac{1}{2}(\sum_{k=1}^n a_k^2)^\frac{1}{2},$$ but I don't know how.
Any help is welcome. Thanks in advance.
Best Answer
Put $A=\left(\sum_{k=1}^n a_k^2\right)^\tfrac{1}{2}$ and $M=\left(2n-\sum_{k=1}^n\frac{1}{k}\right)^\tfrac{1}{2}$. Since we are given $$\sum\limits_{k=1}^n \left(\frac{a_1+a_2+\dots+a_k}{k}\right)\le MA,$$ it suffices to show $$MA-A^2\le \frac{n}{2}-\sum\limits_{k=1}^n\frac{1}{4k}=\frac{M^2}4,$$ which is equivalent to an inequality $$0\le\left(\frac {M}2-A\right)^2.$$