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# 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
Inequalities
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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#### Math Olympiad Program

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