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

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.

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

## 1 Answer

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

• 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? – happymath Feb 21 '17 at 13:02
• @happymath Then edit your question. – Error 404 Feb 21 '17 at 13:03
• @VikrantDesai I am confused whether I should edit or not because there is already an answer – happymath Feb 21 '17 at 13:05
• @happymath You should edit, of course. I will edit my answer accordingly. – 5xum Feb 21 '17 at 13:05