0
$\begingroup$

I just wondered if $$a_1b_1+a_2b_2+a_3b_3+a_4b_4\geq \frac{a_1+a_2+a_3+a_4}{4}(b_1+b_2+b_3+b_4)$$ was true for all positive real numbers $a_i, b_i$ with $1\leq i\leq 4, i \in \mathbb{N}$. I checked some examples and the inequality above held. But on the other hand, I could not find a proof for it. So can anybody find a proof or a counterexample?

$\endgroup$
  • 3
    $\begingroup$ How about $(a_1,\ldots,a_4)=(100,1,1,1)$ and $(b_1,\ldots,b_4)=(1,1,1,100)$? $\endgroup$ – Lord Shark the Unknown May 7 '17 at 11:48
  • $\begingroup$ Ok, this is obviously false. Thank you! $\endgroup$ – mxian May 7 '17 at 11:50
  • $\begingroup$ It is true when $a_i, b_i$ are similarly ordered. Check up on Chebyshev's inequality... $\endgroup$ – Macavity May 7 '17 at 14:31
  • $\begingroup$ You might want to restate it using dot and wedge product of vectors. Let $a=(a_1,a_2,a_3,a_4), b=(b_1,b_2,b_3,b_4), c=(1,1,1,1)$. Then $(c\cdot c)(a\cdot b) - (c\cdot a)(c\cdot b)\ge 0$. Compare this to $(c\wedge a)\cdot(c\wedge b)\ge 0$. In statistics this is related to correlation of sets of data. Similarly ordered data is positively correlated. $\endgroup$ – Somos May 7 '17 at 23:35

Your Answer

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

Browse other questions tagged or ask your own question.