Find the Maximum value of $\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$ if $x$, $y$ and $z$ are positive real numbers such that $x+y+z=4$  Find the maximum value of $$S=\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$$
I tried as follows.
The given expression can be rewritten as
$$S=\sqrt{4-x}+\sqrt{4-y}+\sqrt{4-z}-\left(\frac{y}{\sqrt{x+y}}+\frac{z}{\sqrt{y+z}}+\frac{x}{\sqrt{z+x}}\right)$$
But by symmetry $$S=\left(\frac{y}{\sqrt{x+y}}+\frac{z}{\sqrt{y+z}}+\frac{x}{\sqrt{z+x}}\right)$$
so
$$2S=\sqrt{4-x}+\sqrt{4-y}+\sqrt{4-z}$$  and by Cauchy Scwartz inequality
$$2S \le \sqrt{4-x+4-y+4-z}\times \sqrt{3}$$ so
$$2S \le \sqrt{24}$$
so 
$$S \le \sqrt{6}$$
Is this approach correct?
 A: I think it means that $x$, $y$ and $z$ are non-negatives such that $xy+xz+yz\neq0$.
If $x=3$, $y=1$ and $z=0$ then $S=\frac{5}{2}$.
We'll prove that it's a maximal value.
Indeed, we need to prove that:
$$\sum_{cyc}\frac {x}{\sqrt {x+y}}\leq\frac{5}{4}\sqrt{x+y+z}.$$
By Cauchy-Schwarz 
$$\left(\sum_{cyc}\frac {x}{\sqrt {x+y}}\right)^2\leq\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}.$$
Id est, it remains to prove that
$$\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}\leq\frac{25(x+y+z)}{16}$$ or
$$\sum_{cyc}(8x^6y+72x^6z-14x^5y^2+312x^5z^2-92x^4y^3+74x^4z^3+$$
$$+122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0$$ or 
$$\sum_{cyc}2xy(4x+y)(x-3y)^2(x+2y)^2+$$
$$+\sum_{cyc}(122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0,$$ which is obvious.
Done!
A: No, this approach is not correct, since
$$S=\left(\frac{y}{\sqrt{x+y}}+\frac{z}{\sqrt{y+z}}+\frac{x}{\sqrt{z+x}}\right)$$
does not necessarily hold. For example, consider $x=\frac12$, $y = \frac32$ and $z=2$, then $S=\left(\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}} \right) \approx 2.420$, but $\left(\frac{y}{\sqrt{x+y}}+\frac{z}{\sqrt{y+z}}+\frac{x}{\sqrt{z+x}}\right) \approx 2.445$. 
Also, if this were correct, you would need to provide $x, y$ and $z$ such that $S = \sqrt{6}$. 
A: $$\mathbf{\color{green}{New\ version\ of\ 10.02.2018}}$$
Let us find the greatest value of $\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}$ under the condition $x+y+z=4.$
Substitutions
$$a=\sqrt{x+y},\quad b=\sqrt{y+z},\quad c=\sqrt{z+x}$$
lead to the task on the greatest value of $a+b+c$ under the condition $a^2+b^2+c^2=8.$
Stationary points can be found using Lagrange multipliers method with the function
$$f(a,b,c,\lambda)=a+b+c+\lambda(a^2+b^2+c^2-8)$$
via solution of the system $f'_a=f'_b=f'_c=f'_\lambda=0,$ or
\begin{cases}
1+2\lambda a = 0\\
1+2\lambda b = 0\\
1+2\lambda c=0\\
a^2+b^2+c^2=8\\
(a,b,c)\in \mathbb R_+.
\end{cases}
Summation of $(1.1)-(1.4)$ with factors $a,b,c,-\lambda$ gives
$$a+b+c=-16\lambda,$$ and after substitution of this to $(1)$ one can get the system
$$a(a+b+c)=b(a+b+c)=c(a+b+c),\quad a^2+b^2+c^2=8,$$
with the evident solution
$$a=b=c=\sqrt{\dfrac83},\quad x=y=z=\sqrt{\dfrac43},\quad f=\sqrt{24},$$
so 
$$\boxed{S=\sqrt6\approx2.449490}$$
is the least value.
For example, in the point $(x,y,z)=(0,1,3)$
$$S_1=\dfrac{\sqrt1+\sqrt4+\sqrt3}2=\dfrac{3+\sqrt3}2\approx2.366025 < S.$$
