Inspired by Khang Nguyen Thanh

The criterion and the target function are both homogeneous, and we can therefore additionally assume that $x+y+z = 2$ . This implies that $xy + xz + yz = 1$ . Thus $x,y,z$ are positive roots of the cubic $f(X) = X^3 - 2X^2 + X - q \; = \; 0$ for some positive $q = xyz$ . This cubic has turning points at $X=\tfrac13,1$ , and $f(\tfrac13) = \tfrac{4}{27} - q$ , $f(1) = -q$ . Thus we deduce that $0 < q \le \tfrac{4}{27}$ is the possible range for $q$ .

If $S_j = x^j + y^j + z^j$ then $S_0 = 3$ , $S_1 = 2$ , $S_2 = 2$ and $S_n \; = \; 2S_{n-1} - S_{n-2} + qS_{n-3} \hspace{2cm} n \ge 3$ Thus we calculate $S_3 = 2+3q$ , $S_4 = 2+8q$ and $S_{10} = 2 + 80 q + 175 q^2 + 20 q^3$ , so that we are being asked to maximise $Q(q) \; = \; \frac{2 + 80 q + 175 q^2 + 20 q^3}{4(2+3q)(2+8q)} \hspace{2cm} 0 \le q \le \tfrac{4}{27}$ Since $Q(q) \; =\; \frac{235}{144} + \tfrac{5}{24} q - \frac{277}{180 (2 + 3 q)} - \frac{59}{80 (1 + 4 q)}$ it is easy to see that $Q'(q) > 0$ for $q > 0$ , so that $Q$ is an increasing function of $q>0$ . The maximum value of $Q(q)$ occurs when $q = \tfrac{4}{27}$ , and so the answer is $Q\big(\tfrac{4}{27}\big) = \tfrac{174763}{306504} = \boxed{0.570182}$ .

We know $\sqrt{x}=\sqrt{y}+\sqrt{z}$ by @Steven Jim 's solution.

Let $F=\frac{x^{10}+y^{10}+z^{10}}{(x+y+z)(x^2+y^2+z^2)(x^3+y^3+z^3)(x^4+y^4+z^4)}$

compute that, we get

$F=\frac{(y+z+2\sqrt{yz})^{10}+y^{10}+z^{10}}{(y+z+2\sqrt{yz}+y+z)\big( (y+z+2\sqrt{yz})^2+y^2+z^2 \big) \big( (y+z+2\sqrt{yz})^3+y^3+z^3 \big) \big( (y+z+2\sqrt{yz})^4 +y^4+z^4 \big)}$

Seems that $F$ is a homogeneous term of $t$ , so all divides by $z^{10}$ , we get

$F=\frac{\Big(\frac{y}{z}+1+2\sqrt {\frac{y}{z}}\Big)^{10}+\Big(\frac{y}{z}\Big)^{10}+1}{\prod\limits_{k=1}^{4}\Bigg( \Big(\frac{y}{z}+1+2\sqrt {\frac{y}{z}}\Big)^{k}+\Big(\frac{y}{z}\Big)^k+1)\Bigg)}$

Obviously, $1+2+3+4=10$ . Then, let $t=\frac{y}{z}$

$F(t)=\frac{(\sqrt{t}+1)^{20}+t^{10}+1}{\prod\limits_{k=1}^{4}\Bigg( (\sqrt{t}+1)^{2k}+t^k+1 \Bigg)}$

Up and down both divide $t^{10}$ ,

$F(t)=\frac{\Big( \frac{1}{\sqrt{t}}+1\Big)^{20}+\Big( \frac{1}{t}\Big) ^{10}+1}{\prod\limits_{k=1}^{4}\Bigg( \Big(\frac{1}{\sqrt{t}}+1\Big) ^{2k}+\Big( \frac{1}{t} \Big)^k+1\Bigg)}$

So

$F(t)=F\Bigg( \frac{1}{t}\Bigg)$

$F(1)=0.570181792...$

$F(0.75)=F(1.33)=0.56527487...$

$F(0.25)=F(4)=0.469595654...$

I have seen that $F(1)$ made maximum bound of others $F(t)$ for $0<t<1$ and $t>1$ .

So $Maximum$ occurs when $t=1$ , it mean $y=z$ then $x=4y=4z$

$F(1)=\boxed{0.5702}$