Use the Cauchy-Schwarz Inequality to determine whether $a^2+b^2+c^2$ is bigger than/smaller than/equal to $ab+bc+ac$, where $a,b,c$ are integers and $a<b<c$.
Cauchy-Schwarz Inequality:
$$(\sum_{i=1}^{n}a_ib_i)^2 \leq {\left(\sum_{i=1}^{n}a_i^2\right ) \left ( \sum_{i=1}^{n}b_i^2 \right ) }$$
My attempt:
$n=3$
$a_1=\sqrt{ab}$, $a_2=\sqrt{bc}$, $a_3=\sqrt{ac}$
$b_1=\frac{\sqrt{a}}{\sqrt{b}}$, $b_2=\frac{\sqrt{b}}{\sqrt{c}}$, $b_3=\frac{\sqrt{c}}{\sqrt{a}}$
Plugging it in,
$$ab+bc+ac+\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq a^2 + b^2 + c^2$$
There are $3$ unwanted fractions. Is there any way to remove them?
Best Answer
We should set $a_1=b_3=a, a_2=b_1= b$ and $a_3=b_2=c$ in the Cauchy-Schwarz inequality, to get:
$$(ab+bc+ca)^2\leq (a^2+b^2+c^2)(b^2+c^2+a^2)=(a^2+b^2+c^2)^2$$
and therefore:
$$a^2+b^2+c^2\geq |ab+bc+ca|\geq ab+bc+ca$$
Of course, we don't need any restriction over $a,b,c$ (they don't have to be integers or ordered, they can be any real number).