Prove that $a+b+c\le \frac {a^3}{bc} + \frac {b^3}{ac} + \frac {c^3}{ab}$ One of my friend had just given me an inequality to solve which is stated below. 
Consider the three positive reals $a, b, c$ then prove that 

$$a+b+c\le \frac {a^3}{bc} + \frac {b^3}{ac} + \frac {c^3}{ab}$$

I have solved this inequality very easily using Muirhead. But my friend has no idea what Muirhead inequality is. So I want to know whether there is any other method to solve this problem except for Muirhead's inequality.

Using strictly familiar Angel form of the CS inequality and the traditional CS inequality itself we have: $\displaystyle \sum_{\text{cyclic}} \dfrac{a^3}{bc}= \displaystyle \sum\dfrac{(a^2)^2}{abc}\ge\dfrac{(a^2+b^2+c^2)^2}{abc+abc+abc}=\dfrac{(a^2+b^2+c^2)^2}{3abc}\ge\dfrac{(ab+bc+ca)^2}{3abc}= \dfrac{(ab)^2+(bc)^2+(ca)^2+2a^2bc+2ab^2c+2abc^2}{3abc}\ge\dfrac{(ab)(bc)+(bc)(ca)+(ca)(ab)+2a^2bc+2ab^2c+2abc^2}{3abc}= \dfrac{3a^2bc+3ab^2c+3abc^2}{3abc}=a+b+c$
