Proving Inequality Math Problem for Positive Numbers x, y, and z

[Xamfy19]

hello all:

I have tried to solved the following problem, however, I was stucked. thanks for the help:

x, y z are positive number,
prove (x+y+z)(1/x+1/y+1/z) >=9,

if so, how about (x+y+z+w)(1/x+1/y+1/z+1/w) >= 9.

The following is what I've got:

(x+y+z)(1/x+1/y+1/z) = (x+y+x)[(yz+xz+xy)/xyz]
=[3(xyz) + (x^2*z+x^2*y+y^2*z+x*y^2+y*z^2+x*z^2)]/xyz
= 3 + [(x^2*z+x^2*y+y^2*z+x*y^2+y*z^2+x*z^2)]/xyz

Please help!

 

[dextercioby]

It's no big deal

\left(x+y+z\right)\left(x^{-1}+y^{-1}+z^{-1}\right)=3+\left[\left(xy^{-1}\right)+\left(yx^{-1}\right)\right]+\left[\left(xz^{-1}\right)+\left(zx^{-1}\right)\right]+\left[\left(yz^{-1}\right)+\left(zy^{-1}\right)\right]

\geq 3+2+2+2=9

Q.e.d.

I hope u see why.

Daniel.

 

[dextercioby]

Along the same lines,u can prove quite easily that

\left(x+y+z+w\right)\left(x^{-1}+y^{-1}+z^{-1}+w^{-1}\right)\geq 16

Daniel.

 

[Xamfy19]

Thanks, Daniel;

How did you get the following line?
>= 3 + 2 + 2 +2

Many thanks

 

[dextercioby]

Well,take for example the first

xy^{-1}+yx^{-1} (1)

Make the substitution xy^{-1}=a (2).Then yx^{-1}=a^{-1} (3)

Therefore,(1) becomes a+a^{-1} (4)

And i say that a+a^{-1}\geq 2 (5),for a>0.

Can u prove it...?

Daniel.

 

[Xamfy19]

Thanks, Daniel;

I got it.

 

[Xamfy19]

Cool;

a + 1/a = (a^2 + 1)/a >= 2 since
(a-1)^2 >= 0

 

[dextercioby]

That's right.

Daniel.

 

[primarygun]

Daniel, I prove it with AM-GM Inequalities.

 

[dextercioby]

I think you mean harmonic average.The geometric one is useless...

Daniel.

 

[Vivek Sharma]

arithmatic mean >= harmonic mean
i.e. A>=H
i.e. A/H >= 1

now
. x+y+z+w = 4*A and
1/x + 1/y + 1/z + 1/w = 4/H
hence the product is (4*A)*4/H
=16 * (A/H) >= 16*(1).
QED