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This question already has an answer here:

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.

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marked as duplicate by Martin R, Community Jan 18 '18 at 6:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @Martin R Yep thanks for the link to that question :-) $\endgroup$ – Rohan Shinde Jan 18 '18 at 6:27
  • $\begingroup$ Easy to find with Approach0 :) $\endgroup$ – Martin R Jan 18 '18 at 8:46
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$$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\ge\sqrt{\frac{a^3b^3}{abc^2}}+\sqrt{\frac{b^3c^3}{bca^2}}+\sqrt{\frac{c^3a^3}{cab^2}}=\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge\sqrt{\frac{ab^2c}{ac}}+\sqrt{\frac{bc^2a}{ba}}+\sqrt{\frac{ca^2b}{cb}}=a+b+c$$

Above, we have twice used the known inequality $x^2+y^2+z^2\ge xy+yz+xz$, or (equivalently)$x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{xz}$.

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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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This is the same as $$a^4+b^4+c^4\ge abc(a+b+c).$$ By AM/GM, $$2a^4+b^4+c^4\ge 4a^2bc.$$ Now add the cyclic permutations of this.

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