An inequality with unit coefficients

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]Let $x_1,\ldots ,x_n$ be positive real numbers. Show that there exist $a_1,\ldots ,a_n\in\{-1,1\}$ such that:
\[a_1x_1^2+a_2x_2^2+\ldots +a_nx_n^2\ge (a_1x_1+a_2x_2+\ldots + a_n x_n)^2\]

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Inequalities: An Approach Through Problems
by B.J. Venkatachala
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Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.23.3"]Try using induction.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.23.3"]As the inequality is symmetric in $latex x_1,x_2,\cdots x_n$, introducing an order might help.[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.23.3"]Consider the inequality as $latex f(x_1,x_2,\cdots x_n)\ge 0$.[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.23.3"]

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

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Similar Problems

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