<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] From:
$$\begin{align}
\frac{b+c-a}a&=\frac{(b+c)^2-a^2}{a(a+b+c)}\\
&=\frac{2bc(1+\cos\alpha)}{a(a+b+c)}\\
&=\frac{4b^2c^2\cos^2\frac \alpha2}{abc(a+b+c)}\\
&=\frac{b^2c^2\sin^2\alpha}{abc(a+b+c)\sin^2\frac \alpha2}\\
&=\frac1{\sin^2\frac\alpha2}\frac{4A^2}{abc(a+b+c)}\\
&=\frac1{\sin^2\frac\alpha2}\frac{r}{2R},
\end{align}
$$
where $A,r,R$ are the area , incircle radius and circumcircle radius, respectively, and
recalling that
$$
\frac rR=4\sin\frac\alpha2\sin\frac\beta2\sin\frac\gamma2
$$
we can reformulate the problem as search of extrema of the function
$$
\left(\frac1{\sin x}+\frac1{\sin y}+\frac1{\sin z}\right)\sqrt{2\sin x\sin y\sin z}\tag1
$$
under restrictions 
$$x+y+z=\frac\pi2,\tag2$$
$$z\le y\le x.\tag3$$ 
Here $x=\frac\alpha2$, $y=\frac\beta2$, $z=\frac\gamma2$. The latter restriction is due to the fact that $x$ is the largest angle in the triangle. We assumed without loss of generality $z\le y$.
Fixing the value of $x$ and applying the method of Lagrange multipliers one obtains that the minimum of the function is achieved at
$$
y=z=\frac\pi4-\frac x2\quad\text{ or }\quad\sin y=\sin z=\sqrt{\frac{1-\sin x}2}.\tag4
$$
Observe that both $y$ and $z$ satisfy the restriction $(3)$. Substituting the values into $(1)$ one obtains the expression (i).
As the point $(4)$ is the only critical point of the function $(1)$ its maximum lies on the boundary of the domain. In the case $\frac\pi6\le x<\frac\pi4$ the solution is:
$$
y=x,\; z=\frac\pi2-2x,\quad\text{ or }\quad \sin y=\sin x,\; \sin z=\cos 2x.
$$ 
Substituting the values into $(1)$ one obtains the expression (ii).
If $x\ge\frac\pi4$ there is no positive lower bound for the value of $z$.  It is easy to see that the expression tends to infinity as $z\to0$. This is the statement (iii).

Appendix. Critical points.
To find the local extrema of the function $(1) $ subject to constraint $(2) $ we construct the function:
  $$\left(\frac1{\sin x}+\frac1{\sin y}+\frac1{\sin z}\right)\sqrt{2\sin x\sin y\sin z}-\lambda \left(x+y+z-\frac\pi2\right),\tag {A1}
$$
  where $x$ is assumed to be fixed. To find the extrema of the function (A1) we differentiate it wrt. $y$ and $z $ to obtain:
  $$\begin{cases}
\frac12\frac{\cos y}{\sin y}\left(\frac1{\sin x}-\frac1{\sin y}+\frac1{\sin z}\right)\sqrt{2\sin x\sin y\sin z}=\lambda\\
\frac12\frac{\cos z}{\sin z}\left(\frac1{\sin x}+\frac1{\sin y}-\frac1{\sin z}\right)\sqrt{2\sin x\sin y\sin z}=\lambda
\end{cases}
$$
  which amounts to:
  $$
\frac{\cos y}{\sin y}\left(\frac1{\sin x}-\frac1{\sin y}+\frac1{\sin z}\right)
=\frac{\cos z}{\sin z}\left(\frac1{\sin x}+\frac1{\sin y}-\frac1{\sin z}\right).
$$
  After straightforward algebraic transformation one obtains the equation:
  $$\begin{align}
0&=\sin(y+z)\left(\frac1{\sin z}-\frac1{\sin y}\right)-\sin(y-z)\frac1{\sin x}\\
&=2\sin\frac{y-z}2\left(\frac{\sin(y+z)}{\sin y \sin z}\cos\frac{y+z}2
-\frac1{\sin x}\cos\frac{y-z}2\right)
\end{align}
$$
  The fact that the equation holds for $y=z$ was of course obvious without any derivation. Somewhat harder is to prove that the expression inside the parentheses is strictly positive in
  the whole domain of the variables $y,z$. 
Setting $x=\frac\pi2-y-z$ and reducing to common denominator the claim boils down to
  $$
\forall (y,z)\in{\cal D}_{yz}: 
\sin(y+z)\cos(y+z)\cos\frac{y+z}2-\sin y \sin z \cos\frac{y-z}2>0\tag{A2}
$$
  where ${\cal D}_{yz}$ is the triangle with vertex coordinates $(0,0)$, $\left(\frac\pi4,0\right)$,  $\left(\frac\pi6,\frac\pi6\right)$.
The form of the inequality (A2) suggests the substitution $$u=\frac{y+z}2,  v=\frac{y-z}2,\tag{A3}$$ so that it can be rewritten as:
  $$
\sin4u\cos u +(\cos2u-\cos2v)\cos v>0.\tag{A4}
$$
  Under the linear transform (A3) the domain will be mapped in the following way:
  $$
{\cal D}_{yz}\mapsto {\cal D}_{uv}:
\quad (0,0)\mapsto(0,0),
\; \left(\frac\pi4,0\right)\mapsto\left(\frac\pi8,\frac\pi8\right),
\; \left(\frac\pi6,\frac\pi6\right)\mapsto\left(\frac\pi6,0\right).\tag{A5}
$$ 
  Since $0<\cos v\le 1$ in the considered domain 
  $$\begin{align}
\sin4u\cos u +(\cos2u-\cos2v)\cos v
&\ge \sin4u\cos u +\cos2u-1:=\Phi(u).\tag{A4}
\end{align}
$$
  Differentiating the expression (A4) twice wrt. to $u$ one obtains:
  $$\begin{align}
\frac{\partial^2\Phi(u)}{\partial u^2}
&=-17\sin4u\cos u-8\cos4u\sin u-4\cos 2u\\
&=-9\sin4u\cos u-8\sin5u-4\cos 2u,
\end{align}
$$
  which is negative in the whole domain:
  $$
0<u\le\frac\pi6.
$$
  This means that the function $\Phi(u)$ is concave in the domain. This in turn means that the values of the function in the domain lie above the line defined by the extreme points:
  $$
(u,\Phi(u)):\quad (0,0),
\quad \left(\frac\pi6,\frac14\right),
$$
  which confirms the claim (A2).

</s>