Minimum value of $f(x,y,z) = x^z + y^z - (xy)^{\frac{z}{4}}, x > 0, y > 0, z > 0$

Question

$$f(x,y,z) = x^{z}+y^{z}-(xy)^{\frac{z}{4}}$$ 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?

Answer 1

For a minimum $f_x=zx^{z-1}-(z/4)y^{z/4}x^{z/4-1}=0$ so $x^{3z/4}-y^{z/4}/4=0$.

As $f(x,y,z)=f(y,x,z)$ we also have $y^{3z/4}-x^{z/4}/4=0$ and equating yields $(4x^{3z/4})^3-x^{z/4}/4=0$. Thus $256x^{2z}-1=0$ which gives $x^z=y^z=1/16$ as $x,y,z>0$.

Hence the minimum value is $1/16+1/16-(1/16\cdot1/16)^{1/4}=-1/8$.

Answer 2

An alternative approach: set $a = x^{z/4}$ and $b = y^{z/4}$. Then we want to minimize $a^4 + b^4 - ab$. Due to symmetry, this quantity achieves the minimum value when $a = b$ so we have $2a^4 - a^2$. The derivative is equal to $8a^3 - 2a$ and its equal to $0$ when $4a^2 - 1 = 0$ (since $a > 0$). And we find $a = 1/2$ so the min is $2(1/16) - 1/4 = -1/8$.

Answer 3

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.