Inequality involving arithmetic means: $\prod_{\text{cyc: } a,b,c,d} \frac{a+b}{2} \le \prod_{\text{cyc: } a,b,c,d} \frac{a+b+c}{3}$

Question

Let $a\geq b \geq c\geq d \geq 0$ (ordering matters). Prove that: $$\dfrac{a+b}{2}\dfrac{b+c}{2}\dfrac{c+d}{2}\dfrac{d+a}{2}\leq \dfrac{a+b+c}{3}\dfrac{b+c+d}{3}\dfrac{c+d+a}{3}\dfrac{d+a+b}{3}$$

Note: if $a=b=c=d$, then the result is a direct consequence from property of arithmetic means.

The last fraction should be $\frac{d + a + {\boldsymbol{b}}}{3}$, right? — Caleb Stanford, Sep 16 '15 at 19:41

score 1 · Answer 1 · answered Jun 14 '15 at 11:38

1

This is known as Mongolian inequality and the key tool for proving it is Karamata's inequality.

Full proof (in russian) can be found HERE, page 220.

answered Jun 14 '15 at 11:38

Karlis Olte

637

2

Since the google search for "Mongolian inequality" is heavily interfered with by social inequality in Mongolia, let me add a link to an English proof (complete, I think): http://artofproblemsolving.com/community/c6h26103p163628 – darij grinberg Sep 16 '15 at 21:15

score 0 · Answer 2 · answered Apr 01 '17 at 03:51

0

The Khrabrov's proof see here:

https://artofproblemsolving.com/community/c6h75130p443216

answered Apr 01 '17 at 03:51

Michael Rozenberg

194,933

Inequality involving arithmetic means: $\prod_{\text{cyc: } a,b,c,d} \frac{a+b}{2} \le \prod_{\text{cyc: } a,b,c,d} \frac{a+b+c}{3}$

2 Answers2