Proving Inequality Challenge: $a,\,b,\,c$ | Real Numbers

[anemone]

Let $a,\,b,\,c$ be real numbers such that $a\ge b\ge c>0$.

Prove that $$\frac{a^2-b^2}{c}+\frac{c^2-b^2}{a}+\frac{a^2-c^2}{b}\ge 3a-4b+c$$.

 

[Albert1]

Let $a,\,b,\,c$ be real numbers such that $a\ge b\ge c>0$.

Prove that $$\frac{a^2-b^2}{c}+\frac{c^2-b^2}{a}+\frac{a^2-c^2}{b}\ge 3a-4b+c$$.

Spoiler

$\dfrac{a^2-b^2}{c}+\dfrac{c^2-b^2}{a}+\dfrac{a^2-c^2}{b}\geq \dfrac{a^2-b^2}{c}+\dfrac{a^2-b^2}{a}\\
=\dfrac{(a+c)(a+b)(a-b)}{ac}\geq \dfrac{(a+c)^2(a-b)}{ac}\geq \dfrac{4ac(a-b)}{ac}=4(a-b)\\
\geq 3a-4b+c$

 

[anemone]

Spoiler

$\dfrac{a^2-b^2}{c}+\dfrac{c^2-b^2}{a}+\dfrac{a^2-c^2}{b}\geq \dfrac{a^2-b^2}{c}+\dfrac{a^2-b^2}{a}\\
=\dfrac{(a+c)(a+b)(a-b)}{ac}\geq \dfrac{(a+c)^2(a-b)}{ac}\geq \dfrac{4ac(a-b)}{ac}=4(a-b)\\
\geq 3a-4b+c$

Very well done, Albert! Thanks for participating!(Cool)