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Suppose that $a_1,a_2,...,a_n$ are positive real numbers.Also assume that $b_1,b_2,...b_n$ are an arbitrary permutation of $a_i$'s. Prove that: $$\sum_{i=1}^n a_i^2\geq \sum_{i=1}^n a_ib_i.$$

We know that $\sum\limits_{i=1}^n a_i=\sum\limits_{i=1}^n b_i$ and $\sum\limits_{i=1}^n a_i^2=\sum\limits_{i=1}^nb_i^2$ . I think the rearrangement inequality can't be used to prove it.Maybe some other classic inequality should be used...

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3 Answers 3

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We know that, $\sum_{i=1}^{n}a^2_i=\sum_{i=1}^{n}b^2_i$, then,

$$\sum_{i=1}^{n}(a_i-b_i)^2=\sum_{i=1}^{n}a^2_i+\sum_{i=1}^{n}b^2_i-2\sum_{i=1}^{n}a_ib_i=2\sum_{i=1}^{n}a^2_i-2\sum_{i=1}^{n}a_ib_i\geq 0$$

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    $\begingroup$ Bravo,excellent: simple and nice , +1 and accepted $\endgroup$ Commented Sep 14, 2017 at 3:48
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    $\begingroup$ I agree really elegant +1 $\endgroup$ Commented Sep 14, 2017 at 3:51
  • $\begingroup$ Really smart +1! $\endgroup$
    – MathArt
    Commented Apr 26, 2023 at 7:42
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Note that $$ \sum_{i=1}^na_i^2=\sum_{i=1}^nb_i^2$$ since the $b_i$ are a permutation of the $a_i$. Therefore it follows from the Cauchy-Schwarz inequality that $$ \sum_{i}a_ib_i\leq \Big(\sum_ia_i^2\Big)^{\frac{1}{2}}\Big(\sum_ib_i^2\Big)^{\frac{1}{2}}=\sum_ia_i^2$$

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    $\begingroup$ I didn't see the power of $\frac12$ in CS! $\endgroup$ Commented Sep 14, 2017 at 3:42
  • $\begingroup$ Substituting $a_i$ with $\sqrt{a_i}$, $b_i$ with $\sqrt{b_i}$ in Classical form and both sides taken square roots. \begin{align} \sum_{i=1}^n\sqrt{a_ib_i}\le\sqrt{\sum_{i=1}^n a_i\sum_{i=1}^n b_i} \end{align} Actually for any $\alpha$. $\beta\in\mathbb{R}^+$\begin{align} \biggl(\sum_{i=1}^n x_i^\alpha\,y_i^\beta \biggr)^{\alpha+\beta}\le \biggl(\sum_{i=1}^n x_i^{\alpha+\beta} \biggr)^{\alpha}\biggl(\sum_{i=1}^n y_i^{\alpha+\beta}\biggr)^{\beta}. \end{align} $\endgroup$
    – MathArt
    Commented Apr 26, 2023 at 7:50
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Your inequality it's just the Rearrangement (Chebyshov's) inequality.

Let $a_1\geq a_2\geq...\geq a_n$ and $b_1\geq b_2\geq...\geq b_n$.

Thus, for all $\sigma\in S_n$ we have $$\sum_{i=1}^na_ib_i\geq\sum_{i=1}^na_ib_{\sigma(i)}.$$

Now, take $a_i=b_i$.

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  • $\begingroup$ Thank you,I didn't know about Chebyshov's inequality,(+1) $\endgroup$ Commented Sep 14, 2017 at 4:24
  • $\begingroup$ @Hamid Reza Ebrahimi It's not more inequality, but It's a very useful method. I am ready to show another examples. $\endgroup$ Commented Sep 14, 2017 at 4:25

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