# How to prove $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}+ \frac{3\sqrt[3]{abc}}{a+b+c}\geq 4$ using Middle School methods? [duplicate]

How to prove this:

$$\frac{a}{b} + \frac{b}{c} + \frac{c}{a}+ \frac{3\sqrt[3]{abc}}{a+b+c}\geq 4$$

(Sorry I've asked this question on phone.)

I can understand the C-S step, but I've never learnt about the Schur inequality... I have googled it but I have no idea how to apply it here :/

I've done it by using harmonic mean to the Schur

• are $a,b,c$ assumed to be positive? Oct 5, 2017 at 16:36
• Yes they're. And thanks for editing it! :) Oct 5, 2017 at 16:54
• (i don't think people write "they're" to shorten "they are" in this circumstance) Aug 19, 2020 at 1:43

By C-S $$\sum_{cyc}\frac{a}{b}=\sum_{cyc}\frac{a^2}{ab}\geq\frac{(a+b+c)^2}{ab+ac+bc}.$$ Thus, it remains to prove that $$\frac{(a+b+c)^2}{ab+ac+bc}+\frac{3\sqrt[3]{abc}}{a+b+c}\geq4,$$ which is true and the proof for you.
• @Blate Raven The Schur inequality is the following. For all positives $a$, $b$ and $c$ we have $\sum\limits_{cyc}(a^3-a^2b-a^2c+abc)\geq0.$ Try to use it. Oct 5, 2017 at 19:07