A bound for $\sqrt\frac{b+c-a}a+\sqrt\frac{c+a-b}b+\sqrt\frac{a+b-c}c$ in a triangle Assume that $ABC$ is a triangle with $a\geq b\geq c$, where the angle $A$ has a fixed value. We denote by $\Sigma$ the sum
$$\sqrt\frac{b+c-a}a+\sqrt\frac{c+a-b}b+\sqrt\frac{a+b-c}c.$$
Then the only possible values of $A$ are $\pi/3\leq A<\pi$ and we have:
(i) The smallest possible value $\Sigma$ is
$$\frac{4\sin\frac A2+\sqrt{2\left( 1-\sin\frac A2\right)}}{\sqrt{2\sin\frac A2}}.$$
(ii) If $\pi/3\leq A<\pi/2$ then the largest possible value of $\Sigma$ is
$$\frac{4\cos A+\sqrt{2\left( 1-\cos A\right)}}{\sqrt{2\cos A}}.$$
(iii) If $\pi/2\leq A<\pi$ then there is no finite upper bound for $\Sigma$.   
My question is how to prove (i), (ii), and (iii).
I firstly tried to square the $LHS$, but nothing. I also tried the Radulescu-Maftei theorem, but it didn't help. Hence, I am looking forward to seeing your ideas.

I will prove (i) for now. Use the Ravi substitution:
$$a = y+z,\,\, b = x+z,\,\, c = x+y\implies x\leq y\leq z.$$
And then use the following substitution:
$$\dfrac{2x}{y+z} = X^2,\,\, \dfrac{2y}{x+z} = Y^2,\,\, \dfrac{2z}{x+y} = Z^2\implies X\leq Y\leq Z.$$
After these two, the LHS is simply $X+Y+Z.$ For the RHS, we can easily see:
$$\sin^2\frac A2=\dfrac{1-\cos A}{2} = \dfrac{a^2-b^2-c^2+2bc}{4bc} = \dfrac{(a-b+c)(a+b-c)}{4bc} = \dfrac{yz}{(x+y)(x+z)}=\dfrac{Y^2Z^2}{4}.$$
and so our inequality will be equivalent to:
$$X+Y+Z\geq 2\sqrt{YZ} + \sqrt{\dfrac{2-YZ}{YZ}}.$$
Now I am going to switch to lower-case $X\to x,Y\to y,Z\to z$ for ease of typing. One can check that our $x,y,z$ satisfies:
$$x^2y^2+y^2z^2+z^2x^2 + x^2y^2z^2 = 4$$
and that $yz\geq 1$ because $yz = 2\sin\frac A2\geq 1$ is fixed. 
Finally, we have to prove the following inequality that is equivalent to our original:
$$\sqrt{\dfrac{(2-yz)(2+yz)}{y^2+z^2+y^2z^2}}+y+z\geq 2\sqrt{yz}+\sqrt{\dfrac{2-yz}{yz}}.$$
Subtracting the RHS from the LHS and taking out $(\sqrt{y} - \sqrt{z})^2,$ we are left with:
$$(\sqrt{y} - \sqrt{z})^2\left(1 - \sqrt{\dfrac{2-yz}{yz}}\cdot\dfrac{y+z+2\sqrt{yz}}{\sqrt{yz(2+yz)} + y^2+z^2+y^2z^2}\right)\geq 0,$$
because the two fractions are clearly less than or equal to $1$ due to the fact that $yz\geq 1.$
(iii) is kind of obvious. Take $b = a - \epsilon$ and $c = 2\epsilon.$ Then, the last term on the expression:
$$\dfrac{a+b-c}{c} =\dfrac{2a-3\epsilon}{2\epsilon}\to\infty$$
as $\epsilon\to 0.$
(ii) becomes much more complicated by the same, completing the square approach. Lagrange multiplier works but still more tedious than the nice one in the other answer.
