Let $a_1,...,a_n,b_1,..,b_n$ be integers satisfying $a_1> a_2 \geq a_3 \geq a_4 ... \geq a_n=1$ and $b_1=1$.

Suppose $a_1b_1+....+a_nb_n \geq 0$ then is it true that $a_1^2b_1+...+a_n^2b_n \geq 0$.

My try: I checked this for a few examples and it seems to be true but I don't see how to prove this. Any hints would be appreciated. Thank you.

  • $\begingroup$ Most direct way to prove this is by induction. $\endgroup$ – Paul Feb 21 '17 at 12:59
  • $\begingroup$ This is not true? let $n=2$ and $a_1=10, b_1=-1 b_2=-10$ for instance $\endgroup$ – Jack Yoon Feb 21 '17 at 13:01
  • $\begingroup$ @Paul The statement is false so it cannot be proven by induction. $\endgroup$ – 5xum Feb 21 '17 at 13:02
  • 3
    $\begingroup$ A counterexample with $b_1=1$: $n=3$, $a_1=3$, $a_2=2$, $a_3=1$, $b_1=1$, $b_2=-4$, $b_3=5$. $\endgroup$ – Kelenner Feb 21 '17 at 13:29

Not true for $n=2$, where you can set $a_1=2, a_2=1, b_1=(-1), b_2=2$ and you get

$$2\cdot (-1) + 1\cdot (2)\geq0,$$

however $$2^2\cdot (-1)+1^2\cdot 2 <0.$$

  • $\begingroup$ I actually forgot to put the condition $b_1=1$. So do you still see some counter example? Would you mind if I edited the question to include this? $\endgroup$ – happymath Feb 21 '17 at 13:02
  • $\begingroup$ @happymath Then edit your question. $\endgroup$ – Error 404 Feb 21 '17 at 13:03
  • $\begingroup$ @VikrantDesai I am confused whether I should edit or not because there is already an answer $\endgroup$ – happymath Feb 21 '17 at 13:05
  • 1
    $\begingroup$ @happymath You should edit, of course. I will edit my answer accordingly. $\endgroup$ – 5xum Feb 21 '17 at 13:05

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.