show inequality with Hölder or induction?

Question

For $a_j\in\mathbb R$ with $a_0=0$ show that $\sum_{j=1}^na_j^2\leq n^2\sum_{j=0}^{n-1}(a_{j+1}-a_{j})^2$. First I tried to use induction but this doesn't work. Then I tried to use the Hölder inequality in $\sum_{j=0}^{n-1}(a_{j+1}-a_j)^2=\sum_{j=0}^{n-1}a_{j+1}^2-2\sum_{j=0}^{n-1}a_{j+1}a_j+\sum_{j=0}^{n-1}a_j^2$ but this also doesn't work.

Answer 1

I won't give you the entire solution here, but a hint from where you should be able to finish on your own.

For the sake of convenience, one can define $b_j = a_{j+1} - a_j$ for all $j \in \{0,\ldots, n-1\}$.

Then for any $k \in \{0,\ldots, n\}$, one has

$ a_k = \sum_{j=0}^{k-1}b_j$.

Plugged into the sum expression, this gives

$\sum_{j=1}^{k} a_j^2 = \sum_{j=1}^{n} ( \sum_{k=0}^{j-1} b_k)^2$.

Inductively apply the binomial formula and use

$2ab \leq a^2 + b^2$ for all $a,b \in \mathbb{R}$. Lastly, replace the j in the inner sum with n (upper bound) and use $n (n-1) < n^2$ to finish the proof.