# Question: Using the Cauchy-Schwarz Inequality to compare between 2 expressions

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.

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?

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).

Scalar product:

$$|(u,v)|\le ||u||$$ $$||v||$$.

$$|(a,b,c)\cdot (c,a,b)|\le$$

$$||(a,b,c)||$$ $$||(c,a,b)||$$;

$$ac+ba +bc \le |ac+ba+bc|\le$$

$$a^2+b^2+c^2.$$

$$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(a-b)^2>0.$$

We can get it also, by C-S: $$(1^2+1^2+1^2)(a^2+b^2+c^2)\geq(a+b+c)^2,$$ which is our inequality.