Example $7.18$ Given $a, b, c \geqslant 0, a+b+c=1$, find the maximum value of the following expression.
$$F(a, b, c)=\sqrt{\frac{1-a}{1+a}}+\sqrt{\frac{1-b}{1+b}}+\sqrt{\frac{1-c}{1+c}}$$
The maximum value is $F(0.5,0.5,0)$.
Notice that
$$f^{\prime \prime}=\frac{1-2 x}{\sqrt{(1+x)^{5}(1-x)^{3}}}$$

Assume without loss of generality that $a \leqslant b \leqslant c$.
If $a+b \leqslant \frac{1}{2}$, we have
$$f(a)+f(b) \leqslant f(0)+f(a+b)$$

If $a+b>\frac{1}{2}$, we have
$$f(a)+f(b) \leqslant f\left(a+b-\frac{1}{2}\right)+f\left(\frac{1}{2}\right)$$

Therefore, we can assume without loss of generality that one of $a, b$ is 0 or $\frac{1}{2}$.
If $a=0$, then
$$f(a)+f(b)+f(c) \leqslant 1+\frac{2}{\sqrt{3}}\left(b+c>\frac{1}{2}\right)$$

If $a=\frac{1}{2}, a+b+c \geqslant \frac{3}{2}>1$, which is a contradiction!
If $b=0$, then
$$f(a)+f(b)+f(c)=1+1+0<1+\frac{2}{\sqrt{3}}$$

If $b=\frac{1}{2}$, then
$$f(a)+f(b)+f(c) \leqslant f(0)+f(b)+f(a+c)=1+\frac{2}{\sqrt{3}}$$

Therefore, our assertion is correct.