Special Alex Numbers

Since $x,y,z>0$ , we can divide the second equation by $xyz$ to get:

$\dfrac{1}{z} + \dfrac{9}{y} + \dfrac{25}{x} = 9$ .

Rearranging and using CSI, we have

$\left(\dfrac{5^2}{x} + \dfrac{3^2}{y}+ \dfrac{1^2}{z}\right)(x + y + z) \geq (5+3+1)^2.$

But the LHS also equals $9^2$ according to our givens. The conclusion from CSI must be that corresponding components are proportional, i.e.

$\dfrac{\frac{5^2}{x}}{x} =\dfrac{\frac{3^2}{y}}{y} =\dfrac{\frac{1^2}{z}}{z}$

from which we get the unique solution $x = 5, y = 3, z=1$ .

Hence the sum of all possible values of $xyz$ is $15$ .

xy+9xz+25yz=9xyz; Dividing both sides by xyz, we have that 1/z+9/y+25/x = 9; (z+y+x)(1/z+9/y+25/x) = 81; however, the left hand side of this equation >= ( 1+3+5 )^2 by Cauchy-Schwarz Inequality, which means that the equality must hold; hence we have z/(1/z) = y/(9/y) = x/(25/x); combining this with x+y+z = 9, and noticing that x,y,z are positive, we have that x = 5, y =3, z=1;

Dividing the initial expression for xyz>0, we have

1/z+9/y+25/x=9

And as x+y+z=9, it follows that

(1/z+9/y+25/x)(x+y+z)=81,

where (x/z++25 z/x)+(9 x/y+25 y/x)+(9 z/y+y/z)+1+25+9=81, where

(1) (x/z++25 z/x)+(9 x/y+25 y/x)+(9 z/y+y/z)=46

But the inequality AM-GM, we have

(o) (x/z++25 z/x)>= 2 5=10

(oo) (9 x/y+25 y/x)>= 2*15=30

(ooo) (9*z/y+y/z)>= 2+3=6

Thus (2) (x/z++25 z/x)+(9 x/y+25 y/x)+(9 z/y+y/z)>= 46

with the equal ocorring if, and only if there is equality in (o), (oo) and (ooo).

For equation (1), we see that the equality in (2) must occur, where we have:

• x/z=25*z/x, where x=5z

• 9*z/y=y/z, where y=3z

Thus, 9=x+y+z=5z+3z+z=9z, then z=1 and there is only value xyz, namely:

xyz=(5z) (3z) z=15z^3=15.

Here is a solution inspired from according wiki:

The second equation implies $\frac{1}{z}+\frac{9}{x}+\frac{25}{y}=9.$ Combine with the first equation, we have $z+\frac{1}{z}+x+\frac{9}{x}+y+\frac{25}{y}=18,$ which can be written as $\left(z-2+\frac{1}{z}\right)+\left(x-6+\frac{9}{x}\right)+\left(x-10+\frac{25}{x}\right)=0.$ Since we need only positive solution, we are able to write as $\left(\sqrt{z}-\frac{1}{\sqrt{z}}\right)^2+\left(\sqrt{x}-\frac{3}{\sqrt{x}}\right)^2+\left(\sqrt{y}-\frac{5}{\sqrt{y}}\right)^2=0.$ Hence the only solution is $(x,y,z)=(3,5,1)$ , and the answer is $xyz=\boxed{15}$ .

xy+9xz+25yz=9xyz; Dividing both sides by xyz, we have that 1/z+9/y+25/x = 9; (z+y+x)(1/z+9/y+25/x) = 81; however, the left hand side of this equation >= ( 1+3+5 )^2 by Cauchy-Schwarz Inequality, which means that the equality must hold; hence we have z/(1/z) = y/(9/y) = x/(25/x); combining this with x+y+z = 9, and noticing that x,y,z are positive, we have that x = 5, y =3, z=1;