Prove $5(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2)\geqslant(x_1+x_2+x_3+x_4+x_5)^2$

I can't figure out how to prove

$$5(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2)\geqslant(x_1+x_2+x_3+x_4+x_5)^2$$

Solution 1:

When you expand the right hand side, you get twentyfive terms of the form $x_i x_j$. The left hand side gives twentyfive terms, five each of the form $x_i x_i$.

The equation then reduces to ten sets $x_i x_i + x_j x_j \geq 2 x_i x_j $, and five sets $x_i^2 = x_i^2$. If $x_i = x_j$ then the first set reduces to zero, but in general, we see $x_i^2 + x_j^2 - 2x_i x_j$ is $(x_i-x_j)^2$ which is nonnegative, so the LHS-RHS consists of ten terms for $i \ne j$ which are zero or greater, and five terms $i=j$ which always come to zero.

Thus $n (\sum_{m=1}^n x_m^2) \geq (\sum_{m=1}^n x_m)^2$ for all $n$.

Solution 2:

This is simply Cauchy-Schwarz Inequality. See http://www.artofproblemsolving.com/Wiki/index.php/Cauchy-Schwarz_Inequality

Solution 3:

Have you seen Jensen's Inequality? That would give this inequality in one stroke. A bit more background about where you encountered this problem would help tailor an appropriate answer.