Inequality:

[Fahim Shahriar]

For $a,b,c>0$ , prove that ---
\[\frac{a^3}{bc} + \frac{b^3}{ac} + \frac{c^3}{ab} \geq a+b+c\]

[sakibtanvir]

Its very simple.applying a.m>g.m.we get,
\[\frac{a^3}{bc}+b+c \geq 3a\]
\[\frac{b^3}{ca}+a+c \geq 3b\]
\[\frac{c^3}{ab}+a+b \geq 3c\]
Adding them we get the desired result.

[Phlembac Adib Hasan]

I prefer the proof by AM-GM.(In fact to solve such symmetric inequalities, usually my first choice is AM-GM )
Three other proofs:
$1.$By Holder's Inequality, \[\left (\sum _{cyc}\frac {a^3}{bc}\right )\left (\sum _{cyc}b\right )\left (\sum _{cyc}c\right )\ge (a+b+c)^3\]\[\Longrightarrow \sum _{cyc}\frac {a^3}{bc}\ge a+b+c\]

$2.$WLOG we may assume $a\ge b\ge c$.By Chebyshev's Inequality,\[3\left (\frac {a^3}{bc}+\frac {b^3}{ca}+\frac {c^3}{ab}\right )\ge \left (\frac {a^2}{bc}+\frac {b^2}{ca}+\frac {c^2}{ab}\right )(a+b+c)\ge 3(a+b+c)\]where the last inequality follows from AM-GM.

$3.$By Muirhead's Inequality, \[[3,-1,-1]\ge [1,0,0]\Longleftrightarrow \frac {1}{3}\left (\frac {a^3}{bc}+\frac {b^3}{ca}+\frac {c^3}{ab}\right )\ge \frac {1}{3}\left (a+b+c\right )\]