$\triangle ABC;AB=c;AC=b;BC=a$ such that $a\geq b\geq c$. Prove : $\frac{a^2-b^2}{c}+\frac{b^2-c^2}{a}+\frac{c^2+2a^2}{b}\geq \frac{2ab-2bc+3ca}{b}$

Question

$\triangle ABC;AB=c;AC=b;BC=a$ 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 :

$a\geq b\geq c\Rightarrow \frac{a^{2}-b^{2}}{c}\geq 0;\frac{b^{2}-c^{2}}{a}\geq 0;\frac{3a^{2}}{b}\geq \frac{3ac}{b}$

$\frac{a^2-b^2}{c}+\frac{b^2-c^2}{a}+\frac{c^2+2a^2}{b}\geq \frac{2ab-2bc+3ca}{b}\Leftrightarrow \frac{c^{2}-a^{2}}{b}+\frac{3a^{2}}{b}\geq 2(a-c)+\frac{3ac}{b}\Leftrightarrow \frac{(c-a)(c+a)}{b}\geq 2(a-c)\Leftrightarrow c+a\geq -2b$ !!??

Answer 1

$abc*(LHS-RHS)=-bc^3+ac^3+2abc^2-3a^2c^2+b^3c-2a^2bc+2a^3c-ab^3+a^3b=ab(a^2-b^2)+bc(b^2-c^2)+ac(a-c)(2a-c) \ge 0$

it is trivial when$a=b=c$, the = is hold.$\implies LHS \ge RHS$

Answer 2

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 $a > b > c$.