# Inequality. $\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab$

I want to prove the following inequality :

$$\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab,$$ $a,b,c,d \in \mathbb{R}_{+} .$

In my book, at the answers chapter the author uses AM $\geq$ GM, but I haven't any idea how I can use that.

Thanks :)

• The RHS $= abcd(\sum\frac{1}{a})$ Aug 18, 2012 at 16:13
• The author did't give a complete answer. He suggested that I could resove this inequality using AM $\geq$ GM.
– Iuli
Aug 18, 2012 at 20:38
• @Iuli: what is the book and the name of the author? Aug 18, 2012 at 20:40
• The name of the book is : Inequalities-Theorems-Techniques-and-Selected-Problems and the name of the author is Zdravko-Cvetkovski. If you don't have this book I can send you. This inequality you can find in exercise 113, in the last part of the proof.
– Iuli
Aug 18, 2012 at 20:43
• @Iuli: I know the book and it's worth to have it. Aug 18, 2012 at 21:06

Also notice that:

$$(a+b+c+d)^3 - 16(abc+abd+acd+bcd) = (a+b+c+d)(a+b-c-d)^2 + 4(c-d)^2(a+b) + 4(a-b)^2(c+d) \ge 0$$ Or $$(a+b)[(a+b-c-d)^2 + 4(c-d)^2] + (c+d)[(a+b-c-d)^2 + 4(a-b)^2]\ge0$$

This way is suggested by a friend of mine.

• Please, can you give me a more conclusive answer. Thank :)
– Iuli
Aug 18, 2012 at 16:19
• Ok :) I will wait. Thanks
– Iuli
Aug 18, 2012 at 17:39
• In the first suggestion, terms are either squared terms, and therefore positive, or involve adding together positive numbers, therefore positive. Sep 15, 2012 at 8:30
• This solution is ridiculously nice... (+1) Aug 2, 2013 at 0:45

You need to show that $$(a + b + c + d)^3 - 16(abc + bcd + cda + dab) \geq 0$$ It suffices to show this on any set of the form $0 \leq a,b,c,d \leq N$. By calculus (the "extreme value theorem") the function $(a + b + c + d)^3 - 16(abc + bcd + cda + dab)$ achieves its minimum at some $(a,b,c,d)$ in the set $0 \leq a,b,c,d \leq N$. I claim that this minimum has to occur when $a = b = c = d$.

Write $$(a + b + c + d)^3 - 16(abc + bcd + cda + dab) = (a + b + c + d)^3 - 16(a + b)cd - 16(c + d)ab$$ Note that by AM-GM, we have $16(c + d)ab \leq 16(c + d)({a + b \over 2})^2$. If we had $a \neq b$, we could replace $a$ and $b$ by ${a + b \over 2}$, leaving $c$ and $d$ constant, and we'd get a smaller value. So since $(a,b,c,d)$ is the minimum, this can't happen and we conclude that $a = b$. For similar reasons $c = d$, reversing the roles of the terms $16(a + b)cd$ and $16(c + d)ab$

Next, write $$(a + b + c + d)^3 - 16(abc + bcd + cda + dab) = (a + b + c + d)^3 - 16(b + d)ac - 16 (a + c)bd$$ Then arguing like above gives $b = d$ and $a = c$. Combining the above gives $a = b = c = d$, whereupon $(a + b + c + d)^3 - 16(abc + bcd + cda + dab) = (4a)^3 - 16(4a^3) = 0$. Since this is the minimum, the expression $(a + b + c + d)^3 - 16(abc + bcd + cda + dab)$ is nonnegative as needed.

I posted this inequality on http://www.artofproblemsolving.com/ and I received a nice answer. This answer can be checked on the following link : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=494463 .

• Oh. A brilliant way! It wasn't that hard at all. I went for a while that way, but then I dropped it. I don't remember why. :-) Aug 19, 2012 at 10:43
• Yes, it is a nice solution. Yesterday I told you about that exercise from that book - now I need to find a solution for that inequality using Cauchy-Schwarz. Can you help me, please? I will post the inquality immediately . Thanks:)
– Iuli
Aug 19, 2012 at 11:03

This is a version of Maclaurin's inequality. We have $$S_1=\frac{a+b+c+d}{4}$$ $$S_3=\frac{abc+bcd+cda+dab}{4}$$

and so $$S_1 \geqslant (S_3)^{\frac13}$$ gives the required result.

As noted by Zarrax, it suffices to show that $$(a + b + c + d)^3 - 16(abc + bcd + cda + dab) = 4 \sum_{\mathrm{sym}}a^3 + 36 \sum_{\mathrm{sym}} a^2 b - 40 \sum_{\mathrm{sym}}abc$$ is nonnegative for all $$a, b, c, d \ge 0$$. Here I use the notation $$\sum_{\mathrm{sym}}f(a,b ,c, d) := (1/4!) \sum_{\pi \in S_4} f(\pi(a), \pi(b), \pi(c), \pi(d))$$ as $$\pi$$ ranges over all permutations of $$\{a, b, c, d\}$$.

But by AM-GM, $$a^3 + b^3 + c^3 \ge 3 abc$$ so that $$\sum_{\mathrm{sym}}(a^3 + b^3 + c^3) \ge 3 \sum_{\mathrm{sym}}abc$$ or just $$\sum_{\mathrm{sym}} a^3 \ge \sum_{\mathrm{sym}} abc$$. Similarly, AM-GM also gives $$a^2 b + a^2 c + b^2 a + b^2 c + c^2 a + c^2 b \ge 6 abc$$, so that $$\sum_{\mathrm{sym}} a^2 b \ge \sum_{\mathrm{sym}} abc$$. Thus $$4 \sum_{\mathrm{sym}}a^3 + 36 \sum_{\mathrm{sym}} a^2 b - 40 \sum_{\mathrm{sym}}abc = 4 \left(\sum_{\mathrm{sym}}a^3 - \sum_{\mathrm{sym}}abc\right) + 36 \left(\sum_{\mathrm{sym}} a^2 b - \sum_{\mathrm{sym}}abc\right) \ge 0.$$