Minimum value of $f(x,y,z) = x^z + y^z - (xy)^{\frac{z}{4}}, x > 0, y > 0, z > 0$ $$f(x,y,z) = x^{z}+y^{z}-(xy)^{\frac{z}{4}}$$
for all real positive numbers x, y, z
Does anyone have a clue to find the minimum value of $f(x,y,z)$?
I honestly don't know where to start the solution, I just come up with $AM \geq GM$
$\frac{x^z + y^z}{2} \geq \sqrt{{x}^{z}{y}^{z}} \\ x^z + y^z \geq 2{x}^{\frac{z}{2}}{y}^{\frac{z}{2}}$
With the equality holds if and only if $x^z = y^z$
$x^{z}+y^{z}-(xy)^{\frac{z}{4}} \\ \geq 2{x}^{\frac{z}{2}}{y}^{\frac{z}{2}} - (xy) ^{\frac{z}{4}} \\ = (xy)^{\frac{z}{4}}(2(xy)^{\frac{z}{4}} - 1)$
Set x^z = y^z for minimum value
$(x^{\frac{z}{2}})(2x^{\frac{z}{2}} - 1)$
From here, I set the function $\leq$ 0.
Since x > 0, It's obvious that $x^{\frac{z}{2}}$ can't be $\leq$ 0
$2x^{\frac{z}{2}} - 1 \leq 0 \\ (\sqrt{2} \cdot {x}^{\frac{z}{4}} + 1)(\sqrt{2} \cdot {x}^{\frac{z}{4}} - 1) \leq 0 \\ -\frac{1}{\sqrt{2}} \leq x^{\frac{z}{4}} \leq \frac{1}{\sqrt{2}}$
Since x > 0
$0 < x^{\frac{z}{4}} \leq \frac{1}{\sqrt{2}}$
I don't know what to do after this, I probably did a wrong method to solve the problem. Does anyone have a hint to solve it?

By AM-GM $$x^z+y^z-(xy)^{\frac{z}{4}}\geq2\sqrt{(xy)^z}-(xy)^{\frac{z}{4}}=2\left((xy)^{\frac{z}{4}}-\frac{1}{4}\right)^2-\frac{1}{8}\geq-\frac{1}{8}.$$
The equality occurs for $x^z=y^z=\frac{1}{16},$ which says that we got a minimal value.
