Krdytkyu

Let $x_1,ldots,x_n$ be real numbers such that $|x_1|geq ldotsgeq |x_n|$ and $displaystyle sum_{i=k+1}^n |x_i| leq alpha sum_{i=1}^k |x_i|$ where $1leq k leq n-1$ and $alpha >0$.

Prove that $$sum_{i=k+1}^n x_i^2 leq alpha sum_{i=1}^k x_i^2$$

I have managed to prove a looser inequality: $$begin{align}
alpha sum_{i=1}^k x_i^2
&geq frac{alpha}{k}left(sum_{i=1}^k |x_i| right)^2 quad quad text{Cauchy-Schwarz}\
&geq frac 1k sum_{i=k+1}^n |x_i| sum_{i=1}^k |x_i|\
&geq frac 1k left(sum_{i=k+1}^n |x_i| right)^2\
&geq frac 1k sum_{i=k+1}^n x_i^2 quad quad text{since } |z|_1 geq |z|_2
end{align}$$

Since $frac 1k$ is not $geq 1$, I'm stuck...

It is quite simple: Note that $|x_i| le |x_k|$ for any $i in {k+1,ldots,n}$. Thus
$$tag{1}sum_{i=k+1}^n x_i^2 le |x_k| sum_{i=k+1}^n |x_i| le alpha |x_k| sum_{i=1}^k |x_i|,$$
where the given condition is used in the last step. Now $|x_k x_i| le x_i^2$ for an $i=1,ldots,k$, because $|x_k| le |x_i|$ for all $i=1,ldots,k$. So proceeding in (1) gives the claimed estimate
$$sum_{i=k+1}^n x_i^2lealpha sum_{i=1}^k x_i^2.$$