<s>[INST] prove this inequality $\sum\frac{a}{b}\ge \sum a^2$ Let $a,b,c>0$ such $a+b+c=3$, show that
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge a^2+b^2+c^2$$
I have show this not stronger inequality
$$\sum\dfrac{a}{b}\ge \sum a$$
 [/INST] Let $\sum $ denote cyclic sum, then observe:
$$\sum a(a-b)^2(b-2c)^2 \geqslant 0$$
$$\implies \sum ab^4 + \sum a^3b^2 + 2\sum a^2b^3+4abc\sum ab -8abc\sum a^2 \geqslant 0 $$
$$\implies 2\left( \sum a \right)^2\sum ab^2 + abc\left(\sum a\right)^2 \geqslant 21abc\sum a^2$$
$$\implies 6 \sum \frac{a}b + 3 \geqslant 7\sum a^2$$
Add the obvious $\sum a^2 \geqslant 3$ to the above to conclude. Equality is when $a=b=c=1$.
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