prove this inequality $\sum\frac{a}{b}\ge \sum a^2$

Question

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 $$\sum\dfrac{a}{b}\ge \sum a$$

Answer 1

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 $a=b=c=1$.