Cylic inequality

For $a,b,c$ be positive reals numbers. Proving: 

$$\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a} \geqslant {\frac {3\,{t}^{4}-4\,{t}^{3}-5\,{t}^{2}-2\,t-1}{ \left( t-1 \right) \left( t+1 \right)  \left( 1+2\,t \right) }}.$$

Let $a+b+c=1,ab+bc+ca=\frac{1-t^2}{3}\quad (\,0 \leqslant t \leqslant 1\,).$

We try to find $X$ function such as$:$

$$\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a} \geqslant X$$

Or $$\left( -2\,X-p \right) r+q \left( {p}^{2}-2\,q \right)  \geqslant p \left( a-b \right)  \left( b-c \right)  \left( -c+a \right), $$

Or $$\Big[\left( -2\,X-p \right) r+q \left( {p}^{2}-2\,q \right)\Big]^2 \geqslant p^2 (-4\,r{p}^{3}+{p}^{2}{q}^{2}+18\,pqr-4\,{q}^{3}-27\,{r}^{2})$$

Or $$f(r)=\left( 4{X}^{2}+4Xp+28{p}^{2} \right) {r}^{2}+ \left( -4\,q{p}^ {2}X-20{p}^{3}q+8{q}^{2}X+4p{q}^{2}+4{p}^{5} \right) r+4\,{q}^ {4}\geqslant 0$$

We choose X such as$:$

$$\Delta_r = 16\, \left( {X}^{2}{p}^{2}{q}^{2}-4\,{X}^{2}{q}^{3}-22\,Xp{q}^{3}-2\,X {p}^{5}q+14\,X{p}^{3}{q}^{2}-27\,{q}^{4}+27\,{p}^{4}{q}^{2}+{p}^{8}-10 \,{p}^{2}{q}^{3}-10\,{p}^{6}q \right) {p}^{2} =0,$$

or $$X_{\text{1},\,\, \text{2}}={\frac {11p{q}^{2}-7{p}^{3}q+{p}^{5}\pm 2q(p^2-3q) \sqrt {p^2-3q}}{ \left( {p}^{2}-4\,q \right) q}}$$

We choose $$X = {\frac {11p{q}^{2}-7{p}^{3}q+{p}^{5}+ 2q(p^2-3q) \sqrt {p^2-3q}}{ \left( {p}^{2}-4\,q \right) q}},$$

which means $$X={\frac {3\,{t}^{4}-4\,{t}^{3}-5\,{t}^{2}-2\,t-1}{ \left( t-1 \right) \left( t+1 \right)  \left( 1+2\,t \right) }}.$$

So $$\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a} \geqslant {\frac {3\,{t}^{4}-4\,{t}^{3}-5\,{t}^{2}-2\,t-1}{ \left( t-1 \right) \left( t+1 \right)  \left( 1+2\,t \right) }}.$$

The idea of this is here.