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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?

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3 Answers 3

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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).

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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.$

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

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.

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