A difficult problem

[glyphomouni]

I was trying to prove the following inequality and really got stuck

 

The inequality is. [math]\frac{x^4+y^4+z^4}{x^2y^2+x^2z^2+y^2z^2}\geq\frac{2m+3n}{3(m+n)}[/math] for all the reals x,y,z different to zero and for all the natural Nos m,n

 

Can we use induction here

[Physicsfan]

This post belongs to homework help, and according to forum rules i am not allowed to give you answer. I can only give you hints.

you can express the numerator in terms of [imath](a+b+c)^2[/imath] formula. [imath](a+b+c)^2= a^2+b^2+c^2+2ab+2bc+2ac[/imath].

 

it'l be [imath](a+b+c)^2-2ab-2bc-2ac[/imath]

[imath](a+b+c)^2-2(ab+bc+ac)[/imath]

[imath]=\frac{(a+b+c)^2}{ab+bc+ac}-2[/imath]

 

hey i think you will get L.H.S=0 if you multiply 1/-2 on both sides

Edited June 2, 2010 by Physicsfan

[insane_alien]

This post belongs to homework help

 

that would be why it was posted in the homework help section to begin with.

[the tree]

I'd go for finding the minimum value on the LHS, and maximum on the RHS. Induction may be appropriate for finding the maximum on the right, but probably not really necessary.

[Physicsfan]

that would be why it was posted in the homework help section to begin with.

 

sorry:embarass:

I just want to make a correction LHS wont be zero.

[Mr Skeptic]

I'd see how these numbers compare to the number 1.

[Physicsfan]

i think it is possible to get the LHS as a square and the RHS as a negative number

[the tree]

i think it is possible to get the LHS as a square and the RHS as a negative number

Definitely not. Look at the OP. m and n are to be natural numbers, both sides are strictly positive.

Mr Skeptic is on the right track.

[glyphomouni]

Thank you all ,but could you be more specific?

 

R.H.S is a Natural No ,but what manipulations can make the ratio:[math]\frac{x^4+y^4+z^4}{x^2y^2+x^2z^2+y^2z^2}[/math] such a complicated Natural No??.

 

Why not a whole Natural No??

 

I have to hand in that exercise next Monday and i am still in the dark.

 

Thanx

[the tree]

The LHS isn't a natural number (nor is the RHS), but really that's not the point.

 

Basically, you know that x²,y²,z²,m and n are all strictly positive - that's all you should need.

 

If you can prove that

x²x²+y²y²+z²z²> x²y² + y²z² + z²x²

Then you will know that LHS>1

 

If you can prove that 2m + 3n < 3m + 3n

Then you will know that RHS< 1

 

Since LHS>1>RHS it follows that LHS>RHS.

[glyphomouni]

Thanks .

 

In that case because 2m<3m ,then 2m+3n<3(m+n) or [math]\frac{2m+3n}{3(m+n)}<1\leq\frac{x^4+y^4+z^4}{x^2y^2+x^2z^2+y^2z^2}[/math].

 

I am worried about the equality sign, because [math]A<1\leq B[/math] =>A<B.

 

Is it easy to prove [math]\frac{x^4+y^4+z^4}{x^2y^2+x^2z^2+y^2z^2}\geq 1[/math]??

[the tree]

Is it easy to prove [math]\frac{x^4+y^4+z^4}{x^2y^2+x^2z^2+y^2z^2}\geq 1[/math]??

It's not easy, no, but I managed it in a few lines by taking a geometric view.

 

This does suppose a basic knowledge of dot products, which I'm going to assume because it makes this proof nice and very, very general. (which is always good).

 

Suppose we have two vectors of equal length in one vector space, |v|=|w|.

 

Now it's trivial to prove that:

v.v > v.w

(use the cosine definition of dot products)

 

Now, assign those vectors:

v=(x²,y²,z²)

w=(y²,z²,x²)

 

You'll then have to show that

 

|v|=|w|

 

v.v=x²x²+y²y²+z²z²

 

v.w=x²y²+y²z²+z²x²

 

And then you'll have that

 

x²x²+y²y²+z²z²> x²y² + y²z² + z²x²

 

Which should be all you need.

[glyphomouni]

Before i learn your proof ,there is a thing that i am worried about.

 

if we prove [math]\frac{x^4+y^4+z^4}{x^2y^2+x^2z^2+y^2z^2}\geq 1>\frac{3m+2n}{3(m+n)}[/math] that implies [math]\frac{x^4+y^4+z^4}{x^2y^2+x^2z^2+y^2z^2}>\frac{3m+2n}{3(m+n)}[/math] and not

 

 

[math]\frac{x^4+y^4+z^4}{x^2y^2+x^2z^2+y^2z^2}\geq\frac{3m+2n}{3(m+n)}[/math] .

 

Is that right??

[the tree]

Yes, it is overkill but [ [A>B] => [A> B] ] is so trivial that you shouldn't even need to state it - just state what you've proven, and the conclusion that it implies.

 

(I'm not sure why you'd have been asked to prove that when a slightly more powerful statement is provable, but there you go)

[shyvera]

Another method is to use the Cauchy–Schwarz inequality:

 

[math]\color{white}......[/math][math]x^2y^2+y^2z^2+z^2x^2[/math]

 

[math]\leqslant\ \left(x^4+y^4+z^4\right)^{\frac12}\left(y^4+z^4+x^4\right)^{\frac12}[/math]

 

[math]=\ x^4+y^4+z^4[/math]

 

[math]\therefore\ \frac{x^4+y^4+z^4}{x^2y^2+y^2z^2+z^2x^2}\ \geqslant\ 1\quad\mbox{if}\quad xyz\ne0[/math]

[glyphomouni]

Thank you .

 

But the frame within which we work is high school level. Thus although we know a little bit of vectors we have not done the Cauchy–Schwarz inequality.

 

Anyway thanks for the reference i will look it up