The maximum of squared summation of fixed number of elements

Question

For $n (n\geq1)$ non-negative real numbers $x_1,x_2,\cdots,x_n$, if their summation is fixed $$ \sum_{i=1}^{n}x_i=a, $$ please prove $$ \sum_{i=1}^nx_i^2\leq a^2. $$

My solution:

I prove this question using mathematical induction.

Suppose in the sequence, $m$ elements ($y_1,\cdots y_m$) are positive and the others are zero.

For $m=1$, the result is apparent. For $m=2$, we have $y_1^2+y_2^2\leq(y_1+y_2)^2=a^2$.

Suppose when $m=s (s\geq3)$, the result $$\sum_{i=1}^nx_i^2=\sum_{i=1}^{s}y_i^2\leq a^2$$ holds.

For $m=s+1$,suppose these elements are $y_1,y_2,\cdots y_{s+1}$. We have $$ \sum_{i=1}^{n}x_i^2=\sum_{i=1}^{s+1}y_i^2=y_1^2+y_2^2+\sum_{i=3}^{s}y_i^2\leq(y_1+y_2)^2+\sum_{i=3}^{s}y_i^2\leq a^2. $$

NOTE:

This solution only involves elementary mathematical knowledge, but I want to know whether there is a/an better/advanced solution.

Answer 1

We need to prove that $$\left(\sum_{i=1}^nx_i\right)^2\geq\sum_{i=1}^nx_i^2$$ or $$2\sum_{1\leq i<j\leq n}x_ix_j\geq0,$$ which is obvious.

Also, we can use Karamata for the convex function $f(x)=x^2.$

Indeed, let $x_1\geq x_2\geq...\geq x_n$.

Thus, $(x_1+x_2+...+x_n,0,...,0)\succ(x_1,x_2,...,x_n)$ and our inequality it's just Karamata: $$f(x_1+x_2+...+x_n)+f(0)+...+f(0)\geq f(x_1)+f(x_2)+...+f(x_n).$$