Infimum and supremum of $x^5y^2z$

Question

$A=\{x^5y^2z : x,y,z>0$ and $x+y+z=7\}$

In my mind $\sup A=5^5$ and $\inf A=0$, but how can I prove this?

Answer 1

If $a \in A$ and $a = x^5y^2z; x,y,z > 0$ then $x^5y^2z = a > 0$ so $0 < a$. So $0$ is a lower bound and $0 \le \inf A$.

If $\epsilon > 0$, we can solve $x^5y^2z < \epsilon$ and $x + y + z = 7$ by setting $x = y = \sqrt[7]{\frac \epsilon 7}$ and $z = 7 - 2\sqrt[7]{\frac \epsilon 7}$. (Assuming, of course, that $7 - 2\sqrt[7]{\frac \epsilon 7} > 0$)

So $\epsilon > 0$ can not be a lower bound. So $0$ is the greatest lower bound.

Michael Rozenbergs and Doug M's answer show you that $\frac {5^52^27^8}{8^8}$ is an upper bound.

And $\frac {5^52^27^8}{8} = (\frac {5*7}8)^5(\frac {2*7}8)^2(\frac 78)$ and $\frac {5*7}8 + \frac {2*7}8 + \frac 78 = 7$ we know $\frac {5^52^27^8}{8^8}$ is in $A$ so anything less can not be an upper bound. So $\frac {5^52^27^8}{8^8}$ by definition is the least upper bound.

Answer 2

The hint.

By AM-GM $$x^5y^2z=5^52^2\left(\frac{x}{5}\right)^5\left(\frac{y}{2}\right)^2z\leq5^52^2\left(\frac{5\cdot\frac{x}{5}+2\cdot\frac{y}{2}+z}{8}\right)^8=\frac{5^52^27^8}{8^8}.$$ Now, try to understand that we got a supremum.

Answer 3

$\inf A = 0$ is correct.

You could use something like method Lagrange Mulitipliers to find the upper limit for A.

$f(x,y,z,\lambda) = x^5y^2z - \lambda (x+y+z-7) \frac{\partial f}{\partial x} = 5x^4y^2z - \lambda = 0\\ \frac{\partial f}{\partial y} = 2x^5yz - \lambda = 0\\ \frac{\partial f}{\partial z} = x^5y^2 - \lambda = 0\\ \frac {\lambda x}{5} = \frac {\lambda y}{2} = \lambda z\\ x + \frac {2}{5}x + \frac {1}{5} x = 7\\ x = \frac {35}{8}\\ \sup A = \frac {7^8\cdot5^5}{2^{22}}$

Answer 4

Another equally good method to find the maximum of $A$ is to eliminate one of the three variables. For example, we can eliminate $z$. To this end, write:

$$z = 7 - x - y$$

Substitute this expression into $A$. Now we can find the maximum simply by differentiating the newly found expression for $A$ with respect to the remaning variables, $x$ and $y$. Then set the two derivatives equal to zero. This is a straightforward calculation yielding

$$x^4 y^2 (35 - 6x - 5y) = 0$$

$$x^5 y (14- 2x-3y) = 0$$

Ignoring the pre-factors with the powers of $x$ and $y$, we can solve the set of two linear equations. This way we find $x = 35/8$ and $y = 7/4$. Substituting these values into the expression for $z$ yields $z = 7/8$. Substitution of all three values into $A$ yields the requested result for the maximum value, $A = 5^5 7^8 / 2^{22}$.