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