Nice Ineq

[Nadim Ul Abrar]

Let $a,b,c$ be positive reals so that $ab+bc+ca=abc$
prove that ,
$\displaystyle \frac{a^4+b^4}{ab(a^3+b^3)} +\frac{b^4+c^4}{bc(b^3+c^3)}+\frac{c^4+a^4}{ca(c^3+a^3)} \geq 1$

[*Mahi*]

Hint:

Notice that every term can be manipulated as such:
\[ \frac{a^4+b^4}{a^3+b^3} \geq \sqrt[3]{\frac {a^3+b^3} 2}\], so
\[ \frac{a^4+b^4}{ab(a^3+b^3)} \geq \sqrt[3]{\frac 12 ({\frac 1 {a^3}+ \frac 1 {b^3}}) } \geq \frac 12 (\frac 1a+ \frac 1b)\]
This with the given condition and a little AM-GM finishes the problem.

[Phlembac Adib Hasan]

Let $x=1/a;y=1/b;z=1/c$
The inequality can be re-written as\[\sum _{cyc}\frac{x^4+y^4}{x^3+y^3}\geq 1\]
From AM-GM, $x^4+y^4\ge xy^3+x^3y\Longrightarrow 2(x^4+y^4)\ge (x+y)(x^3+y^3)$
(This result can be obtained also by Chebyshev.)
So it is enough to prove $x+y+z=1$ which is given as a condition.
$ab+bc+ca=abc\Longrightarrow \displaystyle \frac {1}{a}+\frac {1}{b}+\frac {1}{c}=1$