# Proof of AM-GM inequality for $n=3$: $\frac{a+b+c}{3}\geq\sqrt[3]{abc}$ [duplicate]

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Sorry for bad formatting, I couldn't mark the 3rd root on the right hand side... I've figured this out into the point where (and yeah, the problem is to prove that this applies to all non-negative real numbers)

$(a+b+c)/3\geq (abc)^{1/3}$

$$(a+b+c)^3\geq27abc\\ a^3+3a^2b+3a^2c+3ab^2+6abc+3ac^2+b^3+3b^2c+3bc^2+c^3\geq27abc$$

I'm not sure how to proceed. Any advice is appreciated - a full answer would of course be better. Thanks in advance!

## marked as duplicate by Yuriy S, Daniel W. Farlow, JMP, user1551, WatsonJun 28 '16 at 8:24

• This is the arithmetic mean/geometric mean inequality, or AM/GM. Are you doing this on your own, or is it homework? – abiessu Jun 27 '16 at 16:21
• Why are you putting $=$ signs in the beginning of every line? A common misconception is that $=$ means "implies" or "is equivalent to" (it actually means "equals"). You should instead be using $\iff$ in this case (which means "if and only if". Another symbol $\implies$ means "implies"). – user236182 Jun 27 '16 at 16:25
• @user236182: No kiddin'... Read the opening statement ("Sorry for bad formatting, I couldn't mark the 3rd root on the right hand side"). – barak manos Jun 27 '16 at 16:28
• You may be interested in this post of proofs of the AM-GM inequality – Omnomnomnom Jun 27 '16 at 16:33
• See also the proofs from the wiki page – Omnomnomnom Jun 27 '16 at 16:34

If $a,b$ or $c$ equals zero the inequality is trivial. Hence we may assume: $$(a,b,c)=\left(e^{A},e^{B},e^{C}\right)$$ without loss of generality, and by taking $f(x)=e^{x}$ the inequality can be written as: $$\frac{f(A)+f(B)+f(C)}{3}\geq f\left(\frac{A+B+C}{3}\right)$$ that holds as a consequence of the convexity of $f(x)$.

The many online proofs for general $n$ and for $n=2$ leave little to do for $n=3$ except maybe to look for an algebraic identity proving the inequality. Let $(a,b,c) = (x^3,y^3,z^3)$, then:

$x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2 - xy - yz - zx) = (1/2) (x+y+z)(\sum (x-y)^2)$

The conclusion is slightly more precise than the inequality limited to non-negative variables:

the Arithmetic mean minus Geometric mean of $a,b,c$ has the same sign as $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c}$, except for $a=b=c$ where the difference is $0$.

The polynomial in the question, $(a+b+c)^3 - 27abc$, is not factorizable.

For non-negative $a,b,c$ this is provable by elementary methods. For example, for a fixed $c$ and a fixed average of $a$ and $b$ ($\mu = \frac{a+b}{2}$) the triplet is $$(\mu + \delta,\mu-\delta,c)$$ and the largest this can be comes when $\delta = 0$, so the values that make the inequality closest to being false come when $a=b$ and applying that argument again also $a=c$... at which point the equality holds.

But the premise you asked to prove is not restrilcted to non-negative $a,b,c$ and in fact is false: Try $a=-8, b=-4, c=-2$.