How about using Mincopsky's Inequality ? 

By AM-GM's ineq: \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)

\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{\dfrac{c^2+b^2}{2}}+\sqrt{\dfrac{a^2+c^2}{2}}\)

\(\LeftrightarrowΣ_{cyc}\dfrac{a^2}{b}-Σ_{cyc}\sqrt{ab}\geΣ_{cyc}\sqrt{\dfrac{a^2+b^2}{2}}-Σ_{cyc}\sqrt{ab}\left(1\right)\)

We have: \(LHS_{\left(1\right)}\geΣ_{cyc}\dfrac{a^2}{b}-Σ_{cyc}a=Σ_{cyc}\dfrac{\left(a-b\right)^2}{b}\)

And \(RHS_{\left(1\right)}=Σ_{cyc}\dfrac{\left(a-b\right)^2}{2\left(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\right)}\)

Or \(Σ_{cyc}\left(\left(a-b\right)^2\left(\dfrac{1}{b}-\dfrac{1}{2\left(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\right)}\right)\right)\ge0\)

\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\left(\dfrac{\dfrac{\left(\sqrt{2\left(a^2+b^2\right)}+2\sqrt{ab}+b\right)\left(\sqrt{2\left(a^2+b^2\right)}+2\sqrt{ab}-b\right)}{\left(\sqrt{2\left(a^2+b^2\right)}+\sqrt{ab}+b\right)^2}}{\dfrac{1}{b}+\dfrac{1}{2\left(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\right)}}\right)\right)\ge0\)

Done !!

By AM-GM's ineq: \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)

\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{\dfrac{b^2+c^2}{2}}+\sqrt{\dfrac{a^2+c^2}{2}}\)

\(\LeftrightarrowΣ_{cyc}\dfrac{a^2}{b}-Σ_{cyc}\sqrt{ab}\geΣ_{cyc}\sqrt{\dfrac{a^2+b^2}{2}}-Σ_{cyc}\sqrt{ab}\left(1\right)\)

We have: \(LHS_{\left(1\right)}\geΣ_{cyc}\dfrac{a^2}{b}-Σ_{cyc}a=Σ_{cyc}\dfrac{\left(a-b\right)^2}{b}\)

And \(RHS_{\left(1\right)}=\dfrac{\left(a-b\right)^2}{2\left(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\right)}\)

Or \(Σ_{cyc}\left(\left(a-b\right)^2\left(\dfrac{1}{b}-\dfrac{1}{2\left(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\right)}\right)\right)\ge0\)

\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\dfrac{\left(\sqrt{2\left(a^2+b^2\right)}+2\sqrt{ab}+b\right)\left(\sqrt{2\left(a^2+b^2\right)}+2\sqrt{ab}-b\right)}{\dfrac{1}{b}+\dfrac{1}{2\left(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\right)}}\right)\ge0\)

Done !!

By AM-GM's ineq: \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)

\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{\dfrac{b^2+c^2}{2}}+\sqrt{\dfrac{c^2+a^2}{2}}\)

\(\LeftrightarrowΣ_{cyc}\dfrac{a^2}{b}-Σ_{cyc}\left(\sqrt{ab}\right)\geΣ_{cyc}\sqrt{\dfrac{a^2+b^2}{2}}-Σ_{cyc}\left(\sqrt{ab}\right)\left(1\right)\)

We have: \(LHS_{\left(1\right)}\geΣ_{cyc}\dfrac{a^2}{b}-Σ_{cyc}a=Σ_{cyc}\dfrac{\left(a-b\right)^2}{b}\)

And \(RHS_{\left(1\right)}=Σ_{cyc}\left(\dfrac{\left(a-b\right)^2}{2\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\right)\)

Or \(Σ_{cyc}\dfrac{\left(a-b\right)^2}{b}\geΣ_{cyc}\left(\dfrac{\left(a-b\right)^2}{2\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\right)\)

\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\left(\dfrac{1}{b}-\dfrac{1}{2\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\right)\right)\ge0\)

\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\dfrac{\dfrac{\left(\sqrt{2\left(a^2+b^2\right)}+\sqrt{ab}+b\right)\left(\sqrt{2\left(a^2+b^2\right)}+\sqrt{ab}-b\right)}{b^2\left(\sqrt{2\left(a^2+b^2\right)}+\sqrt{ab}\right)^2}}{\dfrac{1}{b}+\dfrac{1}{2\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}}\right)\ge0\)

Done !!!