Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 :)

share|improve this question
    
The RHS $= abcd(\sum\frac{1}{a})$ –  lab bhattacharjee Aug 18 '12 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 '12 at 20:38
    
@Iuli: what is the book and the name of the author? –  Chris's sis Aug 18 '12 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 '12 at 20:43
    
@Iuli: I know the book and it's worth to have it. –  Chris's sis Aug 18 '12 at 21:06
add comment

3 Answers 3

up vote 6 down vote accepted

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.

share|improve this answer
1  
Please, can you give me a more conclusive answer. Thank :) –  Iuli Aug 18 '12 at 16:19
    
Ok :) I will wait. Thanks –  Iuli Aug 18 '12 at 17:39
    
In the first suggestion, terms are either squared terms, and therefore positive, or involve adding together positive numbers, therefore positive. –  Mew Sep 15 '12 at 8:30
    
This solution is ridiculously nice... (+1) –  Prism Aug 2 '13 at 0:45
add comment

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 .

share|improve this answer
    
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. :-) –  Chris's sis Aug 19 '12 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 '12 at 11:03
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.