<s>[INST] prove $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$ 
If $a,b,c$ are positive real numbers and $a+b+c=1$,Prove: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$$
  Additional info:We can use AM-GM and Cauchy inequalities mostly.We are not allowed to use induction.

Things I have done so far:
Using Cauchy inequalities I can write:$$(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a})(b+c+a) \geq (a+b+c)^2$$
$a+b+c=1$,So:$$(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a})\geq 1$$
 [/INST] Homogenization gives $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge \frac{3(a^2+b^2+c^2)}{a+b+c}$$
We'll prove a stronger one $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge 6\cdot \frac{a^2+b^2+c^2}{a+b+c}-(a+b+c)$$
$$\Leftrightarrow (a+b+c)\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\ge 6(a^2+b^2+c^2)-(a+b+c)^2$$
$$\Leftrightarrow (a-b)^2\cdot \frac{a-b+c}{b}+(b-c)^2\cdot \frac{a+b-c}{c}+(c-a)^2\cdot \frac{b+c-a}{a}\ge 0$$
$$\Leftrightarrow S=S_c(a-b)^2+S_a(b-c)^2+S_b(c-a)^2\ge 0$$Since $S_a+S_b=\frac{(a-c)^2+b(a+c)}{ac}\ge 0$
and $S_aS_b+S_bS_c+S_cS_a=\frac{[\sum_{cyc} a^3+3abc-\sum_{cyc} a^2(b+c)]+3abc}{abc}>0$
$$\implies S=\frac{(aS_b+bS_a-cS_a-cS_b)^2+(a-b)^2(S_aS_b+S_bS_c+S_cS_a)}{S_a+S_b}\ge 0$$
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