How do I start this problem? find all real values for $x,y$ and $z$ such that
$x + y + z = 51$
and
$xyz=4000$
given that 
$z\geq 25$ and $0<x\leq 10$
I believe that the only possible solution is $10, 16, 25$ but I do not know how to prove this. I have trying subbing in various ways but I don't think that has helped. The intermixing of inequalities and equalities is something I am not familiar with, so I do not know how to proceed. What kind of techniques are needed?
Any hints to solving this would be appreciated, as I would still like to try solve it myself. If I cannot solve it from those I will reply for more clarification.
 A: The solution by Ahmad can be simplified.
Substituting $z=51-x-y\quad$ we get $\quad xyz=xy(51-x-y)=51xy-x^2y-xy^2=4000$
$y^2+(x-51)y+\frac{4000}x=0\qquad$ 
Assuming there are real solutions, then they are given by $2y=51-x\pm\sqrt{\Delta}\\$ $\text{where }\Delta=(x-51)^2-\frac{16000}x$
Edit: a side study shows $\Delta<0$ when $0<x<1$ so we can assume $x\ge 1$ from now on, it was necessary to justify the statement below (the last implication).
The condition that one of the solution is greater than $25$ implies:
$51-x\pm\sqrt{\Delta}\ge 50\implies \pm\sqrt{\Delta}\ge x-1\implies \Delta\ge (x-1)^2$
Substitution the value for $\Delta$ and using $a^2-b^2$ identity we get:
$(-50)(2x-52)\ge \frac {16000}x\iff 26-x\ge \frac{160}x\iff x^2-26x-160=(x-10)(x-16)\le 0$
Which happens when $x\in[10,16]$.

But since the problem states that $0<x\le 10$ then $x=10$
Now $y^2+(x-51)y+\frac{4000}x=y^2-41y+400=(y-16)(y-25)=0$
Gives the remaining solutions $y=16$ and $z=25$, and we check it verifies the initial problem.
All we had from now where mostly "$\implies$" statements, but it's enough to prove the solution is unique.
A: Solving for $y$ gives $\displaystyle 51-x-z=\frac{4000}{xz}\;\;$, so $\;51xz-x^2z-xz^2=4000\;$ with $0<x\le10$ and $z\ge25$.
Let $f(x,z)=51xz-x^2z-xz^2$ on the rectangular region $0\le x\le10, \;25\le z\le51$ $\hspace{2 in}$(since $z>51\implies y<0\implies xyz<0$);
we will show that $f$ has its maximum value of $4000$ when $x=10, z=25$:
1) On the left side, $g(z)=f(0,z)=0$ for $25\le z\le 51$.
2) On the right side, $g(z)=f(10, z)=410z-10z^2$ is decreasing on $[25,51]$, so its maximum occurs $\hspace{3.4 in}$when $x=10$ and $z=25$.
3) On the top, $h(x)=f(x,51)=-51x^2$ has a maximum value of 0.
4) On the bottom, $h(x)=f(x, 25)=25(26x-x^2)$ is increasing on $[0,10]$, so its maximum occurs $\hspace{3.4 in}$when $x=10$ and $z=25$.
5) Solving $f_{x}=51z-2xz-z^2=0$ and $f_{z}=51x-x^2-2xz=0$ gives the critical points $\hspace{.12 in}(0,0), (51, 0), (0,51), (17, 17)$, and none of these points are in the interior of the region being considered.
Therefore $f(10,25)=4000$ is the maximum value of $f$ on the region $0\le x\le 10, 25\le z\le 51$;
so $x=10, z=25$, and $y=16$ is the only solution.
