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?

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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 $(x,y,z)=(0,1,3)$
$$S_1=\dfrac{\sqrt1+\sqrt4+\sqrt3}2=\dfrac{3+\sqrt3}2\approx2.366025 < S.$$
