Proving AM-GM for the special case $n=3$

Question

I know the AM-GM inequality and its proof which is relatively complex, though the case for $n=2$ is quite simple. However, I don't know of any special easier proof for the case $n=3$, specifically:

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

What is the most elegant proof for this? :)

The proof of the AM-GM inequality is quite straightforward with Cauchy Induction.
The same technique illuminated by mathlove is here used to allow a generalised proof of the AM-GM inequality for $n$ variables.

http://www.artofproblemsolving.com/Wiki/index.php/Cauchy_Induction — Sherlock Holmes, Oct 15 '14 at 00:20

score 6 · Accepted Answer · answered Oct 14 '14 at 17:53

The case for $n=3$ can be proved by using the cases for $n=2,4$.

For $p,q\gt 0$, we have$$(\sqrt p-\sqrt q)^2\ge0\iff \frac{p+q}{2}\ge\sqrt{pq}.$$ So, we have for $s,t,u,v\gt 0,$$$s+t\ge 2\sqrt{st},\ \ \ u+v\ge 2\sqrt{uv}.$$ Hence, we have $$s+t+u+v\ge 2\sqrt{st}+2\sqrt{uv}\ge 2\sqrt{2\sqrt{st}\cdot 2\sqrt{uv}}=4(stuv)^{1/4}.$$

Here, setting $s=a,t=b,u=c,v=\frac{a+b+c}{3}$ gives us $$a+b+c+\frac{a+b+c}{3}\ge 4\left(\frac{abc(a+b+c)}{3}\right)^{1/4}\iff \frac{a+b+c}{3}\ge\sqrt[3]{abc}.$$

neat! i was wondering if the $n=2$ case could be used, but didn't notice that nice little device! — David Holden, Oct 14 '14 at 17:56

score 5 · Answer 2 · answered Oct 14 '14 at 17:50

5

$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = \tfrac12(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2) \ge 0$$

answered Oct 14 '14 at 17:50

Macavity

46,381

This is a far simpler answer than above. – Konstantin Konstantinov Sep 06 '20 at 23:18
@Vladislav What is mentioned is one way of writing and proving 3 variable AM-GM (i.e. $a^3+b^3+c^3\geqslant 3 abc$). To get the OP's version, just replace $x\mapsto x^3$ for each variable in question. – Macavity Oct 12 '23 at 16:51

score 2 · Answer 3 · answered Oct 14 '14 at 17:50

2

Using convexity of $\exp:$

$$ \frac 13(a+b+c) = \frac 13(\exp \log a+\exp \log b+\exp \log c) \\\ge \exp \frac 13( \log a+ \log b+ \log c) = (abc)^{1/3} $$

Bonus: it does not only works for $n=3$ but for any $n$.

answered Oct 14 '14 at 17:50

mookid

28,236

Thanks, though I would have preferred it without calculus :) – user184352 Oct 14 '14 at 17:55
@Bakrala Sen Jensen's is generally considered to not be calculus. For example, it would be generally frowned upon to use Calculus or other nonelementary methods on USAMO, but Jensen's Inequality is perfectly fine. – quasicompactscheme Oct 14 '14 at 18:12
@galois And $\exp$? – user184352 Oct 14 '14 at 18:13
@Bakrala Sen Surely you don't consider $\exp$ to be a nonelementary function. While we do indeed usually define it using methods of calculus, in nonelementary methods the fact that it is an exponential at all is sufficient... We could be using $2^x$ and $\log_2$ and this solution would be no different – quasicompactscheme Oct 14 '14 at 18:14
Alright :) ${}$ – user184352 Oct 14 '14 at 18:17

score 1 · Answer 4 · answered Oct 14 '14 at 17:48

1

since $\ln(x)$ is a concave function on the positive reals, we have, by Jensen's inequality: $$ E(\ln(X)) \le \ln(E(X)) $$ where $X$ is a random variable taking the values $a, b $ or $c$ with equal probability

answered Oct 14 '14 at 17:48

David Holden

18,040

score 0 · Answer 5 · answered Oct 14 '14 at 17:39

powering by three we obtain ${a}^{3}+3\,{a}^{2}b+3\,{a}^{2}c+3\,a{b}^{2}-21\,abc+3\,a{c}^{2}+{b}^{3 }+3\,{b}^{2}c+3\,b{c}^{2}+{c}^{3} \geq 0$ with $b=a+u,c=a+v$ we get $\left( 9\,{u}^{2}-9\,uv+9\,{v}^{2} \right) a+{u}^{3}+3\,{u}^{2}v+3\,u {v}^{2}+{v}^{3} \geq 0$ which is true.

score 0 · Answer 6 · answered Oct 14 '14 at 19:37

0

If we can use other inequalities, by Rearrangement $$a.a.a+b.b.b+c.c.c \ge a.b.c+b.c.a+c.a.b$$ generalises.

answered Oct 14 '14 at 19:37

Macavity

46,381

Proving AM-GM for the special case $n=3$

6 Answers6