Valentine won't show it merci

By completing the squares, we can rewrite the initial condition as $\left(x + y - \dfrac{3}{2}\right)^2 + \left(z - \dfrac{3}{2}\right)^2 = \dfrac{9}{2}$ .

Let $m = x+y+z$ . I will now prove a lemma, namely that $m \le 6$ .

Let's use Jensen's Inequality on the function $y = x^2$ : $\dfrac{9}{2} = \left(x + y - \dfrac{3}{2}\right)^2 + \left(z - \dfrac{3}{2}\right)^2$ $\dfrac{9}{2} \ge 2\left(\dfrac{1}{2}\left(x + y - \dfrac{3}{2} + z - \dfrac{3}{2}\right)\right)^2 = \dfrac{(x+y+z-3)^2}{2}$ $x+y+z-3 \le 3$ $x+y+z \le 6. \boxed{}$

Now let's use Jensen's Inequality on the function $y = \dfrac{20}{\sqrt{x}}$ : $x + y + z + \dfrac{20}{\sqrt{x+z}} + \dfrac{20}{\sqrt{y+2}} \ge x + y + z + 2*\dfrac{20}{\sqrt{\dfrac{x+y+z+2}{2}}}$ $= x+y+z + \dfrac{40\sqrt{2}}{\sqrt{x+y+z+2}} = m + \dfrac{40\sqrt{2}}{\sqrt{m + 2}}$

We now need one more little lemma: that the function $f(m) = m + \dfrac{40\sqrt{2}}{\sqrt{m + 2}}$ is decreasing for $0 \le m \le 6$ : $f'(m) = 1 - \dfrac{20\sqrt{2}}{(m+2)^{3/2}} \le 1 - \dfrac{20\sqrt{2}}{(6+2)^{3/2}} = -\dfrac{1}{4} < 0$

Therefore, the minimum of $f(m)$ for $0 \le m \le 6$ occurs at $m = 6$ , and it is $f(6) = 6 + \dfrac{40\sqrt{2}}{\sqrt{6 + 2}} = 26$

Therefore, $x + y + z + \dfrac{20}{\sqrt{x+z}} + \dfrac{20}{\sqrt{y+2}} \ge 26$ . This value is achieved when $x = 1, y = 2, z = 3$ . Hence the minimum value is $\boxed{26}$ .

Because student in grade 10 in the competition only-can using AM-GM and C-S(The other must prove then can use).I post a solution using only AM-GM.

From the conditional we have: $3(x+y+z)=(x+y)^2+z^2\geq \frac{1}{2}(x+y+z)^2$ $\Rightarrow x+y+z\leq 6$ Rewrite the polynomial: $[(x+z)+\frac{8}{\sqrt{x+z}}+\frac{8}{\sqrt{x+z}}]+[(y+2)+\frac{8}{\sqrt{y+2}}+\frac{8}{\sqrt{y+2}}]+4[\frac{1}{\sqrt{x+y}}+\frac{1}{\sqrt{y+2}}]-2$ Call the expression above is P.By applying AM-GM we have: $P\geq 3\sqrt[3]{(x+z)\frac{64}{x+z}}+3\sqrt[3]{(y+2)\frac{64}{y+2}}+\frac{8}{\sqrt[4]{(x+z)(y+2)}}-2$ $=22+\frac{8\sqrt{2}}{\sqrt{x+y+z+2}}\geq 26$ So $max=26$ .The equality holds when $x=1,y=2,z=3$

@Ariel Gershon and @ASWIN T.S. both of you have good solution.Thanks.

since the answer is not in decimal

x + z is a square number

y+ 2 is a square number

to get a minimum value for the expression, x+z and y+2 should be small square number

therefore y=2

and x=1 and z=3

therefore value of expression = 1 +2+3 +20/2 +20/2

                                           = 1+2+3+20

                                           = 26