Understand the problem

Let $x_1,\ldots ,x_n$ be positive real numbers. Show that there exist $a_1,\ldots ,a_n\in\{-1,1\}$ such that:
\[a_1x_1^2+a_2x_2^2+\ldots +a_nx_n^2\ge (a_1x_1+a_2x_2+\ldots + a_n x_n)^2\]

Source of the problem
Iberoamerican olympiad 2011
Topic
Inequalities
Difficulty Level
Easy
Suggested Book
Inequalities: An Approach Through Problems
by B.J. Venkatachala

Start with hints

Do you really need a hint? Try it first!

Try using induction.
As the inequality is symmetric in x_1,x_2,\cdots x_n, introducing an order might help.
Consider the inequality as f(x_1,x_2,\cdots x_n)\ge 0.

Let us assume that x_1\ge x_2\ge\cdots\ge x_n.   Claim a_{odd}=1, a_{even}=-1 works.   Proof: For n=1 Trivial.   For n=2 The inequality is equivalent to x_1^2-x_2^2\ge (x_1-x_2)^2. Expanding, this becomes x_1x_2\ge x_2^2 which is a consequence of x_1\ge x_2.   For n\ge 3   Consider a_1x_1^2+a_2x_2^2+\cdots +a_nx_n^2-(a_1x_1+a_2x_2+\cdots +a_nx_n)^2 as a function of x_1 (say f(x_1)). We need to show that the minimum of f is at least 0. Clearly, f is linear in x_1 and the coefficient of x_1 is 2(x_2-x_3+x_4-x_5\cdots +(-1)^nx_n). Due to the order chosen, this coefficient is non-negative. Hence the minimum is attained at the minimum of x_1, which is x_2. However, putting x_1=x_2, we are left with a_3x_3^2+a_4x_4^2+\cdots +a_nx_n^2\ge (a_3x_3+a_4x_4+\cdots a_nx_n)^2 which is true from the induction hypothesis.   QED

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