Decimate the decimals

I'm typing this out at 3:11 AM cause I can't sleep. If there;s any errors please just tell:

Lemma: $z^n+y^n\le \frac{2}{2^{2n}}\left(\sqrt{y}+\sqrt{z}\right)^{2n}$ for all real $n$ Proof: Since we need to prove for all real $n$ , it indicates that we should use a continuous method, namely calculus $z^n + y^n = \left(\sqrt{y}+\sqrt{z}\right)^{2n}\cdot \frac{1+\left(\frac{z}{y}\right)^n}{\left(1+\sqrt{\frac{z}{y}}\right)^{2n}}$ Using calculus, it can be found that $\frac{1+k^n}{\left(1+\sqrt{k}\right)^{2n}}=\frac{1+\left(\frac{z}{y}\right)^n}{\left(1+\sqrt{\frac{z}{y}}\right)^{2n}}\le \frac{2}{2^{2n}}$ The inequality follows from here.

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First thing to notice is that $x^2 + y^2 + z^2 = 2(xy+yz+zx) = 2$ .

$x^2 + y^2 + z^2 - 2(xy + yz + zx) = (\sqrt{x} + \sqrt{y} + \sqrt{z})(-\sqrt{x} + \sqrt{y} + \sqrt{z})(\sqrt{x} - \sqrt{y} + \sqrt{z})(\sqrt{x} + \sqrt{y} - \sqrt{z})=0$

Without loss of generality, we can assume $x>y \ge z$ . Hence $\sqrt{x} = \sqrt{y} + \sqrt{z}$

$x^n + y^n + z^n = \left( \sqrt{y} + \sqrt{z} \right)^{2n} + y^n + z^n$

Now since $z^n+y^n\le \frac{2}{2^{2n}}\left(\sqrt{y}+\sqrt{z}\right)^{2n}$

$\left( \sqrt{y} + \sqrt{z} \right)^{2n} + y^n + z^n \le \left(1+\frac{2}{2^{2n}}\right)\left(\sqrt{y}+\sqrt{z}\right)^{2n}=\left(1+\frac{2}{2^{2n}}\right)\cdot x^n$

Now since $z+y=2-x,\ yz=1-zx-xy=1-x(2-x)$ and $\left(y+z\right)^2\ge 4yz$ through AM-GM, $\max_x=\frac{4}{3}$ .

Hence $x^n + y^n + z^n \le \left(1+\frac{2}{2^{2n}}\right)\cdot x^n \le \left(1+\frac{2}{2^{2n}}\right)\cdot \left(\frac{4}{3}\right)^n$ With equality achieved when $4y=4z=x=\frac{4}{3}$