# Small inequality

Proposition: Let $x_1,x_2,x_3$ and $a_1,a_2,a_3$ be real numbers and $x_1+x_2+x_3=0$. Prove that $a_1x_1+a_2x_2+a_3x_3=0$.

My solution:

Case 1: If $x_1,x_2,x_3\geq0$, then

$a_1x_1+a_2x_2+a_3x_3\leq\max\{|a_1|,|a_2|,|a_3|\}(x_1+x_2+x_3)=0$

But in general, is the proposition true?

• Where did you find that problem? In my opinion, it doesn't make sense since the fact you have to prove is not true in general Commented May 7, 2015 at 3:12
• I don't think so. If $a_1x_1+a_2x_2+a_3x-3 = 0$ why is $(a_1+1)x_1+... =0$? Commented May 7, 2015 at 3:13
• It is not true. Let $x_1=1$, $x_2=1$, $x_3=-2$. Let $a_1=2$, $a_2=1$, $a_3=1$. Of course if the $x_i$ are $\ge 0$, then $x_1+x_2+x_3=0$ forces all the $x_i$ to be $0$, and then $\sum a_ix_i=0$. Commented May 7, 2015 at 3:14

The proposition in general is not true. It is only true if $x_1=x_2=x_3 = 0$. Because if $a_1x_1 + a_2x_2 + a_3x_3 = 0$, $(a_1+1)x_1 + a_2x_2 + a_3x_3$ is not always 0 (except for the special case)and $a_1+1$ is obviously real.