Is This Inequality Solvable for Positive Variables?

[evagelos]

can some body help me with the following inequality.



...x/(y+z) +y/(z+x) + z/(x+y)>=3/2......

...for x>0,y>0,z>0.........

 

[snipez90]

Hmm I know about 8 proofs for this inequality. The simplest ones still require a transformation. Without giving away too much, can you rewrite the left-hand side so that each term looks more "similar"? Anyways the algebraic manipulation is the key. Once you've made it you can pretty much apply the mean chain of inequalities (AM-GM is the most common one), normalize the inequality, or various other methods.

 

[evagelos]

8 ? I can't even make one.
please show me all.is there a book for impossible inequalities having a general plan for
solving them??
please inform me

 

[snipez90]

Ok maybe the last hint wasn't enough. On the left hand side, can you make each term's numerator the same? It shouldn't be hard to determine what that expression for the numerators should be if you consider what the denominators look like (y+z, z+x, x+y).

Anyways, this inequality is meant to demonstrate various methods of solving even harder inequalities. You won't learn anything if I showed you the proofs and you don't know some basics about inequalities to begin with. Good inequality material usually stems from good problem solving books such as Problem Solving Strategies by Arthur Engel. But the least you should know is the Arithmetic Mean - Geometric Mean inequality and the Cauchy-Schwarz inequality. Then there are probably about a dozen more well-known and relatively elementary inequalities. After that it's just using those inequalities and algebraic manipulation.

 

[evagelos]

man i know the stuff ,H<=G<=A,CSI and the rest but inequalities are always a puzzle to
me.
Thanks for the tip so we must prove that :

(x+y+z){ 1/(x+y) + 1/(x+z) + 1/(z+y)}>=9/2 using H<=A
Now is there any other way apart from this one ??
i tell you what ,i will post another difficult inequality ,so please look at the new post
and help me

 

[snipez90]

Well could you prove it using AM-HM? You're very close. Can you make the two products on the left hand side look more similar? Multiply both sides by 2 and then rewrite the (x+y+z) term to match the reciprocals then it immediately follows from AM-HM.

 

[evagelos]

.thanks iknow how to do it now but please go to another thread for inequalities again
and please if you can help me with the new one .i posted just now under general maths again



.

 

[evagelos]

suppose somebody cannot find the trick to add 3 to both sides of the inequality and then transform it into the proper shape so to use Hm--Am,and instead of that does all the
calculations and cancellations and ends up with the following equivalent inequality:

2x^3 + 2y^3 + 2z^3 >= yx^2 + xy^2 + zx^2 + xz^2 + zy^2 + yz^2.


how can we proceed from here to solve the inequality??

 

[snipez90]

$$(x-y)^2(x+y) \geq 0$$

 

[evagelos]

.........very good............
now,where did you get the idea to transform :

...x^3 + y^3 - yx^2 -xy^2 into (x +y)(x-y)^2?

 

[snipez90]

Well the first thing I tried was $$(x^2 + y^2)(x+y)$$. I had the $$x^2$$ and the $$x$$ parts already since I needed an $$x^3$$ term. Obviously this was not correct since it resulted in $$x^3 + x^2y + xy^2 + y^3$$. I needed the two middle terms to be negative to prove the inequality. Hence looking back again I realized if I changed the signs in the products to negatives, the "first" and "last" resulting terms would be positive while the "middle" two would be negative.

 

[Kurret]

Also try the very fundamental "rearrangement inequality".
http://en.wikipedia.org/wiki/Rearrangement_inequality
And consider x,y,z to be one sequence of numbers, and 1/(x+y), 1/(x+z),1/(y+z) be the other sequence of numbers.

 

[evagelos]

snipez90 either you are a magician or you hiding something.

Thanks anyway for your help to put my mind in the correct path of the puzzle