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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 :)

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  • $\begingroup$ The RHS $= abcd(\sum\frac{1}{a})$ $\endgroup$ Aug 18, 2012 at 16:13
  • $\begingroup$ The author did't give a complete answer. He suggested that I could resove this inequality using AM $\geq$ GM. $\endgroup$
    – Iuli
    Aug 18, 2012 at 20:38
  • $\begingroup$ @Iuli: what is the book and the name of the author? $\endgroup$ Aug 18, 2012 at 20:40
  • $\begingroup$ 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. $\endgroup$
    – Iuli
    Aug 18, 2012 at 20:43
  • $\begingroup$ @Iuli: I know the book and it's worth to have it. $\endgroup$ Aug 18, 2012 at 21:06

5 Answers 5

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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.

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

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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 .

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  • $\begingroup$ 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. :-) $\endgroup$ Aug 19, 2012 at 10:43
  • $\begingroup$ 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:) $\endgroup$
    – Iuli
    Aug 19, 2012 at 11:03
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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.

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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 $$4 \sum_{\mathrm{sym}}a^3 + 36 \sum_{\mathrm{sym}} a^2 b - 40 \sum_{\mathrm{sym}}abc = 4 \left(\sum_{\mathrm{sym}}a^3 - \sum_{\mathrm{sym}}abc\right) + 36 \left(\sum_{\mathrm{sym}} a^2 b - \sum_{\mathrm{sym}}abc\right) \ge 0.$$

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