$a,b,c > 0$ (no other conditions)
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\sqrt{3\left(a^2+b^2+c^2\right)}$
I tried this: $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{\left (a+b+c\right)^2}{a+b+c}=a+b+c$ then $a+b+c\geq\sqrt{3\left(a^2+b^2+c^2\right)}$ which is not correct.
As the inequality is homogeneous, WLOG we can set $a^2+b^2+c^2 = 3$ to get the equivalent inequality: $$\frac{a^2}b + \frac{b^2}c + \frac{c^2}a \ge 3$$
$$\iff \left(\frac{a^2}b + \frac{b^2}c + \frac{c^2}a \right)^2 \ge 9 \iff \sum_{cyc} \frac{a^4}{b^2} \ge 3 \iff \sum_{cyc} a^2 \left(\frac{a^2}{b^2}+\frac{b^2}{a^2}-2 \right)\ge 0$$
Another way, using Holder's Inequality: $$\left(\frac{a^2}b + \frac{b^2}c + \frac{c^2}a \right)\left(\frac{a^2}b + \frac{b^2}c + \frac{c^2}a \right)\left(a^2b^2 + b^2c^2 + c^2a^2 \right)\ge \left(a^2+b^2+c^2\right)^3$$
Let $x=a^2, y=b^2, z=c^2$. Then it remains sufficient to show that $$ \left(x+y+z \right)^3 \ge 3\left(xy+yz+zx \right)(x+y+z)$$ $$ \iff \left(x+y+z \right)^2 \ge 3\left(xy+yz+zx \right)$$ which is easy to show.
if $$a^2\to a,b^2\to b,c^2\to c$$ then $a+b+c=3$,we have $$\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{c}}+\dfrac{c}{\sqrt{a}}\ge 3$$ so you can see this Inequality. $\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3$
