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
$$5(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2)\geqslant(x_1+x_2+x_3+x_4+x_5)^2$$
 A: 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 $n$.  
A: This is simply Cauchy-Schwarz Inequality. See http://www.artofproblemsolving.com/Wiki/index.php/Cauchy-Schwarz_Inequality
A: 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.
A: Here's a proof by induction. Let $s_n:=n\sum_{i=1}^nx_i^2$ and $t_n:=\sum_{i=1}^nx_i\;(i=1,...,n).$ We will prove that $u_n:=s_n-t_n^2\geqslant0\;(n\in\Bbb N).$ First, the result holds for $n=0$: $s_0-t_0^2=0-0=0\geqslant0.$ Now suppose the result to hold for $n=k,$ where for convenience we will write $x$ for $x_{k+1}$:$$u_{k+1}=(k+1)\left(\dfrac{1}{k}s_k+x^2\right)-(t_k+x)^2=kx^2-2t_kx+\dfrac{k+1}{k}s_k-t_k^2.$$This quadratic form in $x$ has its minimum at $x=t_k/k$, where $$u_{k+1}=\dfrac{1}{k}t_k^2-\dfrac{2}{k}t_k^2+\dfrac{k+1}{k}s_k-t_k^2=\dfrac{k+1}{k}(s_k-t_k^2)\geqslant0,$$completing the inductive step.
A: $$5({x_1}^2+{x_2}^2+{x_3}^2+{x_4}^2+{x_5}^2)\geqslant(x_1+x_2+x_3+x_4+x_5)^2.$$
\begin{align}
&\text{let } |x_i|=x_i^+. \\
\ \\
&\Rightarrow \bigg|\sum x_i\bigg| \leqslant \sum x_i^+. \\
\ \\
&\Rightarrow \bigg(\sum x_i\bigg)^2 \leqslant \bigg( \sum x_i^+ \bigg)^2 \\
&\leqslant5\bigg(\sum x_i^{+^2}\bigg) \Big( \text{Using Cauchy-Schwarz inequality.} \Big) \\
& =5\bigg(\sum {x_i}^2\bigg).
\end{align}
