Calvin says why be unequal if you can be equal?

Letting $a=\dfrac{x+y}{x+y+z}$ , $b=\dfrac{y+z}{x+y+z}$ , and $c=\dfrac{z+x}{x+y+z}$ we see that $a+b+c=2$ and we want to maximize $\sqrt{a}+\sqrt{b}+\sqrt{c}$

But note that by Cauchy Schwarz, $(\sqrt{a}+\sqrt{b}+\sqrt{c})^2\le 3(a+b+c)=6$ so $\sqrt{a}+\sqrt{b}+\sqrt{c}\le \sqrt{6}\approx \boxed{2.45}$

Here is my solution:

Let $a=\frac{x+y}{x+y+z},b=\frac{y+z}{x+y+z},c=\frac{x+z}{x+y+z}$ . So, $a+b+c=2$ and we want to maximize $\sqrt a+\sqrt b+\sqrt c$ . So, let $f(x)=\sqrt x$ . Since $f(x)$ is a concave function, by Jensen's Inequality, we have $f(\frac{a+b+c}{3}) \leq \frac{f(a)+f(b)+f(c)}{3}$ this implies that $f(a)+f(b)+f(c) \leq \sqrt 6$ , and thus, the maximum value is $\sqrt 6=\boxed{2.45}$

Let, $a=\sqrt{\dfrac{x+y}{x+y+z}}, \,b=\sqrt{\dfrac{y+z}{x+y+z}}$ , and $c=\sqrt{\dfrac{z+x}{x+y+z}}$

$\begin{aligned}&\implies a^2+b^2+c^2=2\\ &\implies \frac{1}{2}[(a-b)^2+(b-c)^2+(c-a)^2]=2-ab-bc-ca \quad \color{#3D99F6}[\text{subtracting } (ab+bc+ca)\text{ from both sides}]\\ &\text{But, } (a-b)^2+(b-c)^2+(c-a)^2\geq 0 \quad \color{#3D99F6}\text{[for real } a,b,c]\\ &\implies 2-ab-bc-ca\geq 0\\ &\implies ab+bc+ca\leq 2\\ &\text{Now, }(a+b+c)^2=a^2+b^2+ c^2+2(ab+bc+ca)\\ &\implies (a+b+c)^2\leq(2+2\times 2)=6\\ &\implies \boxed{a+b+c \leq \sqrt{6}\approx 2.45}\end{aligned}$

$\color{#D61F06}\text{NB:} \color{#3D99F6} \text{ This method also shows that } (a+b+c) \text{ is maximum when } \\ \color{#20A900}(a-b)^2+(b-c)^2+(c-a)^2=0\implies a=b=c\implies x=y=z$ .

let x+y+z = a then we need to maximize: (a-z/a)^0.5 +(a-x/a)^0.5 +(a-y/a)^0.5 = b by rms>+am we get 6^0.5>=b