1. ## a long equation

Let x, y and z be three positive real numbers bigger than 0, where xyz=6. Prove that:
2x/((2x^2+y^2)(x^2+2z^2))+3y/((3y^2+z^2)(y^2+3x^2))+5z/((5z^2+x^2)(z^2+5y^2)) is smaller or equal to 1/8.

2. ## Re: a long equation

To rewrite to make it easier to read, you are asking for help proving that:

$x,y,z\in \mathbb{R}^+, xyz = 6 \Longrightarrow \dfrac{2x}{(2x^2+y^2)(x^2+2z^2)}+\dfrac{3y}{(3y^2+ z^2)(y^2+3x^2)}+\dfrac{5z}{(5z^2+x^2)(z^2+5y^2)} \le \dfrac{1}{8}$

Let's multiply out. Since xyz is positive, each factor in the denominator is positive, so we can multiply out by everything in the denominator and we will not change the direction of the inequality.

This is the equivalent of saying:

$x,y,z \in \mathbb{R}^+, xyz = 6 \Longrightarrow 2x(3y^2+z^2)(y^2+3x^2)(5z^2+x^2)(z^2+5y^2)+3y(2x^2 +y^2)(x^2+2z^2)(5z^2+x^2)(z^2+5y^2)+5z(2x^2+y^2)(x ^2+2z^2)(3y^2+z^2)(y^2+3x^2) \le \dfrac{(2x^2+y^2)(x^2+2z^2)(3y^2+z^2)(y^2+3x^2)(5z ^2+x^2)(z^2+5y^2)}{8}$

Expanding out the LHS and RHS (using wolframalpha):
$8(30 x^6 y^3 + 90 x^6 y^2 z + 6 x^6 y z^2 + 30 x^6 z^3 + 90 x^5 y^4 + 48 x^5 y^2 z^2 + 6 x^5 z^4 + 15 x^4 y^5 + 75 x^4 y^4 z + 213 x^4 y^3 z^2 + 205 x^4 y^2 z^3 + 42 x^4 y z^4 + 60 x^4 z^5 + 30 x^3 y^6 + 466 x^3 y^4 z^2 + 242 x^3 y^2 z^4 + 30 x^3 z^6 + 15 x^2 y^6 z + 105 x^2 y^5 z^2 + 155 x^2 y^4 z^3 + 321 x^2 y^3 z^4 + 50 x^2 y^2 z^5 + 60 x^2 y z^6 + 150 x y^6 z^2 + 80 x y^4 z^4 + 10 x y^2 z^6 + 30 y^6 z^3 + 150 y^5 z^4 + 10 y^4 z^5 + 30 y^3 z^6) \le 90 x^8 y^4 + 48 x^8 y^2 z^2 + 6 x^8 z^4 + 75 x^6 y^6 + 670 x^6 y^4 z^2 + 341 x^6 y^2 z^4 + 42 x^6 z^6 + 15 x^4 y^8 + 533 x^4 y^6 z^2 + 1181 x^4 y^4 z^4 + 515 x^4 y^2 z^6 + 60 x^4 z^8 + 105 x^2 y^8 z^2 + 806 x^2 y^6 z^4 + 407 x^2 y^4 z^6 + 50 x^2 y^2 z^8 + 150 y^8 z^4 + 80 y^6 z^6 + 10 y^4 z^8$

What have you done so far?

3. ## Re: a long equation

Nothing, actually. And I'm pretty sure this problem isn't meant to be solved by expanding the RHS and LHS.

4. ## Re: a long equation

Originally Posted by slipeternal
expanding out the lhs and rhs (using wolframalpha):
$8(30 x^6 y^3 + 90 x^6 y^2 z + 6 x^6 y z^2 + 30 x^6 z^3 + 90 x^5 y^4 + 48 x^5 y^2 z^2 + 6 x^5 z^4 + 15 x^4 y^5 + 75 x^4 y^4 z + 213 x^4 y^3 z^2 + 205 x^4 y^2 z^3 + 42 x^4 y z^4 + 60 x^4 z^5 + 30 x^3 y^6 + 466 x^3 y^4 z^2 + 242 x^3 y^2 z^4 + 30 x^3 z^6 + 15 x^2 y^6 z + 105 x^2 y^5 z^2 + 155 x^2 y^4 z^3 + 321 x^2 y^3 z^4 + 50 x^2 y^2 z^5 + 60 x^2 y z^6 + 150 x y^6 z^2 + 80 x y^4 z^4 + 10 x y^2 z^6 + 30 y^6 z^3 + 150 y^5 z^4 + 10 y^4 z^5 + 30 y^3 z^6) \le 90 x^8 y^4 + 48 x^8 y^2 z^2 + 6 x^8 z^4 + 75 x^6 y^6 + 670 x^6 y^4 z^2 + 341 x^6 y^2 z^4 + 42 x^6 z^6 + 15 x^4 y^8 + 533 x^4 y^6 z^2 + 1181 x^4 y^4 z^4 + 515 x^4 y^2 z^6 + 60 x^4 z^8 + 105 x^2 y^8 z^2 + 806 x^2 y^6 z^4 + 407 x^2 y^4 z^6 + 50 x^2 y^2 z^8 + 150 y^8 z^4 + 80 y^6 z^6 + 10 y^4 z^8$
yikes!!

5. ## Re: a long equation

Originally Posted by louis33
Nothing, actually. And I'm pretty sure this problem isn't meant to be solved by expanding the RHS and LHS.
Then my guess is you can somehow maximize each term individually and add them together. According to Wolframalpha, each term individually has a maximum of $\dfrac{1}{24}$. There is probably a way to determine that fairly easily. Then, you know the sum has a maximum that is no greater than the sum of the maxima of each term individually which would give the $\dfrac{1}{8}$. If you give it a try and still get stuck, I will think about it a little more.

6. ## Re: a long equation

If x, y, and z are "three positive real numbers bigger than 0, where xyz=6" then either x= 1, y= 2, z= 3 (or some permutation) or x= 1, y= 1, z= 6 (or some permutation)! That gives a total of 6+ 3= 9 possible combinations,
x= 1, y= 2, z= 3
x= 1, y= 3, z= 2
x= 2, y= 1, z= 3
x= 2, y= 3, z= 2
x= 3, y= 1, z= 2
x= 3, y= 2, z= 1
x= 1, y= 1, z= 6
x= 1, y= 6, z= 1
x= 6, y= 1, z= 1
Calculate 2x/((2x^2+y^2)(x^2+2z^2))+3y/((3y^2+z^2)(y^2+3x^2))+5z/((5z^2+x^2)(z^2+5y^2)) for each of those 9 cases.