question
Let the real numbers $a,b,c$:
a) If $a^2+ab+bc+ca<0$, show that $a^2<b^2+c^2$
b) If $b^2+c^2+ab+bc+ca<0$, show that $a^2>b^2+c^2$.
idea
The first thing that came into my mind is the fact that we can demonstrate by AM-GM that
$a^2+b^2+c^2\geq ab+bc+ca$
which basically means that
$b^2+c^2\geq ab+bc+ca-a^2$
Maybe we can decrease this inequality from $a^2+ab+bc+ca<0$.
I don't know what to do forward. Hope one of you can help me! Thank you!
Best Answer
Using $(a+b+c)^2\ge 0$ helps as commented by Calvin Lin and Bob Dobbs.
a) $$\begin{align}a^2&\le a^2+\underbrace{(a+b+c)^2}_{\ge 0} \\\\&=2a^2+b^2+c^2+2ab+2bc+2ca \\\\&=b^2+c^2+\underbrace{2(a^2+ab+bc+ca)}_{\lt 0} \\\\&\lt b^2+c^2\end{align}$$
b) $$\begin{align}b^2+c^2&\le b^2+c^2+\underbrace{(a+b+c)^2}_{\ge 0} \\\\&=a^2+2b^2+2c^2+2ab+2bc+2ca \\\\&=a^2+\underbrace{2(b^2+c^2+ab+bc+ca)}_{\lt 0} \\\\&\lt a^2\end{align}$$