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.

Prove : $$\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq\left(\frac{a+b}{2}\right)^3+\left(\frac{c+d}{2}\right)^3$$ For $a,b,c,d>0$

share|improve this question
6  
Why do you ask? What have you tried? –  Did Aug 23 '11 at 11:54
1  
It should be solvable using bunching. en.wikipedia.org/wiki/Muirhead%27s_inequality#Examples –  David Schwartz Aug 23 '11 at 12:49
1  
@David: if I wanted to do that, the denominators in the LHS and the lack of symmetry of both sides would annoy me. But maybe you know how to circumvent these... –  Did Aug 23 '11 at 13:02
2  
I'm tring to solve it use holder inequality ،then i've got : $$\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq\left‌​(\frac{a+d}{2}\right)^3+\left(\frac{b+c}{2}\right)^3$$ ..is it usefull ? –  Lina Aug 23 '11 at 13:07
1  
@Lina: Kudos for showing your work. This reasoning could succeed if $(a+d)^3+(b+c)^3\ge(a+b)^3+(c+d)^3$ but there is no reason to believe this last inequality holds (and it happens to be false in general). –  Did Aug 23 '11 at 13:19
show 2 more comments

1 Answer 1

up vote 4 down vote accepted

I know many time has passed, however i came up with a solution and I hope it is correct: here you are:

Let us denote $$\begin{array}{c}\frac{1}{4}\frac{(b+c)^3}{2}(8(a^2+d^2)^3-(a+d)^3(a+b)^3)+\\\frac{1}{4}\frac{(a+d)^3}{2}(8(b^2+c^2)^3-(c+d)^3(c+b)^3)=(\clubsuit)\end{array}$$

And we want to prove $(\clubsuit)\geq0.$

We recall $x^3-y^3=(x-y)(x^2+xy+y^2)$ $(\spadesuit)$ and so $$\begin{array}{c}(\clubsuit)=(2(a^2+d^2)-(a+d)(a+b))C_1+2((b^2+c^2)-(c+d)(c+b))C_2\geq\\ \min\{C_1,C_2\}\cdot(2(a^2+b^2+c^2+d^2)-(a+d)(a+b)-(c+d)(c+d)).\end{array}$$

Where $C_1,C_2$ are nonnegative costants depending on $a,b,c,d$, and can be easily derived following $(\spadesuit)$

But now it is true that $$2(a^2+b^2+c^2+d^2)-a^2-ab-ad-bd-c^2-cd-cb-bd\geq 0,$$ since $$\begin{eqnarray} b^2+d^2&\geq& 2bd,\\\frac{a^2+d^2}{2}&\geq& ad,\\ \frac{c^2+d^2}{2}&\geq& cd,\\\frac{c^2+b^2}{2}&\geq&cb,\\\frac{a^2+b^2}{2}&\geq&ab,\\a^2&\geq&a^2,\\c^2&\geq&c^2.\end{eqnarray}$$

We have then estabilished $(\clubsuit)\geq 0$. But it is easy to show that this relation implies $$(a^2+d^2)^3(b+c)^3+(b^2+c^2)^3(a+d)^3\geq\frac{(a+b)^3(a+d)^3(b+c)^3}{8}+\frac{(c+d)^3(a+d)^3(b+c)^3}{8}$$ which in turn implies $$\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq \left(\frac{a+b}{2}\right)^3+\left(\frac{c+d}{2}\right)^3,$$ as desired.

share|improve this answer
    
Proof by Texas Hold'em :) –  user13838 Sep 15 '11 at 15:14
1  
$\clubsuit$, $\heartsuit$, $\spadesuit$, $\diamondsuit$ are the best symbols in Latex. They combine two of my great passions: maths and Hold'em :) –  uforoboa Sep 15 '11 at 15:53
    
Thank you very much –  Lina Sep 16 '11 at 14:17
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.