<s>[INST] 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.
 [/INST] (i) With$$\frac{\pi}{3}\le A <\frac{\pi}{2}$$let $\angle B=\angle C$

In the isosceles triangle$$\sqrt{\frac{b+c-a}{a}}=\sqrt{\frac{b+c}{a}-1}=\sqrt{\csc \frac{A}{2}-1}$$Again$$\sqrt{\frac{c+a-b}{b}}=\sqrt{2\sin \frac{A}{2}}$$and also$$\sqrt{\frac{a+b-c}{c}}=\sqrt{2\sin \frac{A}{2}}$$Thus$$\Sigma=2\sqrt{2\sin \frac{A}{2}}+\sqrt{\frac{1-\sin \frac{A}{2}}{\sin \frac{A}{2}}}$$And putting over a common denominator and multiplying top and bottom by $\sqrt2$
$$\Sigma=\frac{4\sin \frac{A}{2}+\sqrt{2}\cdot\sqrt{1-\sin \frac{A}{2}}}{\sqrt2\cdot \sqrt{\sin \frac{A}{2}}}=\frac{4\sin \frac{A}{2}+\sqrt{2(1-\sin \frac{A}{2})}}{\sqrt{2\sin \frac{A}{2}}}$$the proposed expression.
(ii) In the next figure, suppose$$\frac{\pi}{3}\le B=A<\frac{\pi}{2}$$
Then$$\sqrt\frac{b+c-a}{a}=\sqrt\frac{c}{a}=\sqrt{2\cos A}$$and$$\sqrt\frac{c+a-b}{b}=\sqrt\frac{c}{b}=\sqrt{2\cos A}$$and$$\sqrt\frac{a+b-c}{c}=\sqrt{\frac{a+b}{c}-1}=\sqrt{\sec A-1}$$Hence$$\Sigma=2\sqrt{2\cos A}+\sqrt{\sec A-1}=2\sqrt{2\cos A}+\sqrt{\frac{1-\cos A}{\cos A}}$$Putting over a common denominator$$\Sigma=\frac{2\sqrt2\cdot\sqrt{\cos A}\cdot\sqrt{\cos A}+\sqrt{1-\cos A}}{\sqrt{\cos A}}$$And multiplying top and bottom by $\sqrt2$,$$\Sigma=\frac{4\cos A+\sqrt2\cdot\sqrt{1-\cos A}}{\sqrt{2\cos A}}=\frac{4\cos A+\sqrt{2(1-\cos A)}}{\sqrt{2\cos A}}$$Thus for given$$\frac{\pi}{3}\le A<\frac{\pi}{2}$$we get the other proposed expression when $B=A$.
To examine for maximum and minimum, note the limited range of positions for $B$.
Building on the first figure above, we see that for given $\angle A$, point $B$ is confined between $B'$ and $B''$, beyond which the condition that$$A\ge B\ge C$$ does not hold. 
Further, since$$\triangle AB'C\sim \triangle AB''C$$then the three quantities$$\sqrt{\frac{b+c-a}{a}}, \sqrt{\frac{c+a-b}{b}}, \sqrt{\frac{a+b-c}{c}}$$ are the same, respectively, in these two limiting triangles, and hence the extreme value of $\Sigma$ is the same. 
To see that it is a maximum, note that if $A=\frac{\pi}{3}$, then points $B'$ and $B''$ coincide with $B$, and since $\sin \frac{\pi}{6}=\cos \frac{\pi}{3}$, then$$\frac{4\sin \frac{A}{2}+\sqrt{2(1-\sin \frac{A}{2})}}{\sqrt{2\sin \frac{A}{2}}}=\frac{4\cos A+\sqrt{2(1-\cos A)}}{\sqrt{2\cos A}}=3$$Hence no distinction between maximum and minimum. On the other hand, if $A=\frac{\pi}{2}$, and is thus no longer acute, then$$\frac{4\sin \frac{A}{2}+\sqrt{2(1-\sin \frac{A}{2})}}{\sqrt{2\sin \frac{A}{2}}}\approx3.022$$while$$\frac{4\cos A+\sqrt{2(1-\cos A)}}{\sqrt{2\cos A}}=\frac{0+\sqrt2}{0}$$has passed all bounds.
Hence abbreviating the left sides as $f(\sin \frac{A}{2})$ and $f(\cos A)$, it is clear that, for $\frac{\pi}{3}<A<\frac{\pi}{2}$, the ratio$$\frac{f(\cos A)}{f(\sin \frac{A}{2})}$$grows beyond all bounds with increasing $A$. 
$\Sigma$ is thus a maximum in (ii), when $\angle A=\angle B$ or $\angle C$, and a minimum in (i), when $\angle B=\angle C$.
(iii) Finally for$$\frac{\pi}{2}\le A<\pi$$
With $\angle A$ and length $c$ fixed, if we move point $C$ toward point $A$, then 
$\sqrt{\frac{a+b-c}{c}}$ and $\sqrt{\frac{b+c-a}{a}}$ each approaches zero, while $\sqrt{\frac{a+c-b}{b}}$ exceeds all bounds.  Thus $\Sigma$ is unbounded in this case.
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