$\triangle ABC;AB=c;AC=b;BC=a$ such that $a\geq b\geq c$. Prove : $\frac{a^2-b^2}{c}+\frac{b^2-c^2}{a}+\frac{c^2+2a^2}{b}\geq \frac{2ab-2bc+3ca}{b}$ $\triangle ABC;AB=c;AC=b;BC=a$ such that $a\geq b\geq c$. Prove :
$$\frac{a^2-b^2}{c}+\frac{b^2-c^2}{a}+\frac{c^2+2a^2}{b}\geq \frac{2ab-2bc+3ca}{b}$$
I have tried that :
$a\geq b\geq c\Rightarrow \frac{a^{2}-b^{2}}{c}\geq 0;\frac{b^{2}-c^{2}}{a}\geq 0;\frac{3a^{2}}{b}\geq \frac{3ac}{b}$ 
$\frac{a^2-b^2}{c}+\frac{b^2-c^2}{a}+\frac{c^2+2a^2}{b}\geq \frac{2ab-2bc+3ca}{b}\Leftrightarrow \frac{c^{2}-a^{2}}{b}+\frac{3a^{2}}{b}\geq 2(a-c)+\frac{3ac}{b}\Leftrightarrow \frac{(c-a)(c+a)}{b}\geq 2(a-c)\Leftrightarrow c+a\geq -2b$ 
!!??

Another way to look at it would be:
\begin{align}
LHS &= \frac{(a - b)(a + b)}{c} + \frac{(b - c)(b + c)}{a} + \frac{c^2 + 2a^2}{b} \\
&> (a - b) + (b - c) + \frac{c^2 + 2a^2}{b} \\
&= a - c + \frac{c^2 + 2a^2}{b}. 
\end{align}
And 
RHS $= 2a - 2c + \dfrac{3ca}{b}$.
\begin{align}
\text{So, }LHS > RHS &\iff \frac{c^2 + 2a^2 - 3ac}{b} > a - c \\
&\iff c^2 + 2a^2 - 3ac > ab - bc \\
&\iff (a - c)^2 + a(a - c) > b(a - c)  \\
&\iff a - c + a > b \\
&\iff 2a > b + c
\end{align}
and this last inequality is true since $a > b > c$.
