1
$\begingroup$

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...

$\endgroup$
3
$\begingroup$

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

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

$\endgroup$
  • 1
    $\begingroup$ Bravo,excellent: simple and nice , +1 and accepted $\endgroup$ – Hamid Reza Ebrahimi Sep 14 '17 at 3:48
  • 1
    $\begingroup$ I agree really elegant +1 $\endgroup$ – Isham Sep 14 '17 at 3:51
1
$\begingroup$

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 $$ \sum_{i}a_ib_i\leq \Big(\sum_ia_i^2\Big)^{\frac{1}{2}}\Big(\sum_ib_i^2\Big)^{\frac{1}{2}}=\sum_ia_i^2$$

$\endgroup$
  • 1
    $\begingroup$ I didn't see the power of $\frac12$ in CS! $\endgroup$ – Hamid Reza Ebrahimi Sep 14 '17 at 3:42
1
$\begingroup$

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 $a_i=b_i$.

$\endgroup$
  • $\begingroup$ Thank you,I didn't know about Chebyshov's inequality,(+1) $\endgroup$ – Hamid Reza Ebrahimi Sep 14 '17 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. $\endgroup$ – Michael Rozenberg Sep 14 '17 at 4:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.