# 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.

• each answer looks like it comes from quadratic formula – mathworker21 Dec 1 '19 at 23:14
• Use the Ravi substitution and then make $\dfrac{x}{y+z}\to x', \dfrac{y}{z+x}\to y'$ and $\dfrac{z}{y+x}\to z'$. Once you have done this, you will get a constraint, quadratic optimization problem without any square root. – dezdichado Dec 2 '19 at 2:04
• Why can't $A$ be smaller than $\pi/3$? – YiFan Dec 2 '19 at 3:33
• @YiFan : Because the side opposite $A$ (of length $a$) is the longest. – John Bentin Dec 2 '19 at 13:44
• @John: thanks, I thought one just chooses the labels $a,b,c$ based on the lengths of the sides. – YiFan Dec 2 '19 at 14:55

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: $$0 This means that the function $$\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).

• a very nice way to solve this in a somewhat standard way ! However, could you give me more details on how to solve the system we obtain when we apply the method of Lagrange Multipliers? The gradient looks pretty ugly and I don't know if that system can be solved without the aid of a CAS. How did you do it? – JoMath Dec 2 '19 at 18:02
• The application of the Lagrange method in this case is not complicated. It appeared however much harder to prove that the resulting equation has no other solutions except for $y=z$. I added now such a proof, which can be probably drastically simplified. – user Dec 3 '19 at 14:53

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.

• very nice and quite elementary I would say ! I am really looking forward to seeing your solution to (ii) tomorrow ! – JoMath Dec 2 '19 at 17:49
• @JoMath, elementary method turned out to be too ugly to continue. Lagrange Multiplier with two constraint function worked out in the end but it was longer and more tedious than the other answer though. – dezdichado Dec 3 '19 at 22:10
• The problem is not anymoreactive, but I"ll wait for the new issue of GMA . after that, I'll post my solution. Very nice approach. The second solution too. As for the 3-rd... – user120123 Dec 4 '19 at 10:52
• @user120123 what is GMA? is the problem from there – dezdichado Dec 5 '19 at 19:39
• @dezdichado, the problem was not anymore active at the date of posting.So it's alright. GMA=GAzeta Matematica Seria A, for college students and professors (even teachers).I'm the author and I found it here by a happy accident.I'm glad I did See 485 from ssmr.ro/gazeta/gma/2018/gma3-4-2018-continut.pdf – user120123 Dec 6 '19 at 22:37

(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 $$\frac{\pi}{3}, the ratio$$\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.

• Thank you! Your idea is nice and intuitive, but it is not really rigorous. – JoMath Dec 5 '19 at 19:32
• @JoMath-I've elaborated a bit, in order to make the maximum/minimum argument more convincing. – Edward Porcella Dec 5 '19 at 20:10

Author's solution (mine): Since $$a\geq b\geq c$$ we have $$A\geq B\geq C$$. Since $$A$$ is the largest angle of a triangle, we have $$\pi/3\leq A<\pi$$.

Let $$x=\frac{b+c-a}a$$, $$y=\frac{c+a-b}b$$, $$z=\frac{a+b-c}c$$, so that $$\Sigma =x+y+z$$. We have the well known relations: $$x=\frac{2\sin\frac B2\sin\frac C2}{\sin\frac A2},\quad y=\frac{2\sin\frac C2\sin\frac A2}{\sin\frac B2},\quad z=\frac{2\sin\frac A2\sin\frac B2}{\sin\frac C2}.$$ It follows that $$yz=4\sin^2\frac A2,\quad zx=4\sin^2\frac B2,\quad xy=4\sin^2\frac C2\quad\text{and}\quad xyz=8\sin\frac A2\sin\frac B2\sin\frac C2.$$ Hence the identity $$\sin^2\frac A2+\sin^2\frac B2+\sin^2\frac C2+2\sin\frac A2\sin\frac B2\sin\frac C2=1$$ writes as $$xy+yz+zx+xyz=4$$.

Since $$\frac\pi 2>\frac A2\geq\frac B2\geq\frac C2>0$$ we have $$yz\geq zx\geq xy$$ so $$x\leq y\leq z$$.

Let $$k=yz=4\sin^2\frac A2$$. Since $$A$$ is fixed, so is $$k$$. We have $$\pi/3\leq A<\pi$$ so $$4\sin^2\frac\pi 6\leq k<4\sin^2\frac\pi 2$$, i.e. $$1\leq k<4$$.

Let $$y+z=2s$$. We also have $$yz=k$$ so, by the AM-GM inequality, $$s\geq\sqrt k$$.

We have $$x(y+z+yz)=4-yz$$, i.e. $$x(2s+k)=4-k$$, so $$x=\frac{4-k}{2s+k}$$. We also have $$\sqrt y+\sqrt z=\sqrt{y+z+2\sqrt{yz}}=\sqrt{2s+2\sqrt k}$$. Hence $$\Sigma =\sqrt{2s+2\sqrt k}+\sqrt\frac{4-k}{2s+k}=f_k(s),$$ where $$f_k:[\sqrt k,\infty )\to{\mathbb R}$$, $$f_k(t)=\sqrt{2t+2\sqrt k}+\sqrt\frac{4-k}{2t+k}.$$

We claim that $$f_k$$ is strictly increasing. We have $$f'_k(t)=\frac 1{\sqrt{2t+2\sqrt k}}-\sqrt\frac{4-k}{(2t+k)^3}$$. We must prove that $$f'_k(t)>0$$, i.e. $$(2t+k)^3>2(t+\sqrt k)(4-k)$$, $$\forall t\geq\sqrt k$$. Since $$k\geq 1$$ we have $$(2t+k)^3\geq (2t+1)^3$$ and $$3\geq 4-k$$. Hence it suffices to prove that $$(2t+1)^3>6(t+\sqrt k)$$, i.e. $$8t^3+12t^2+6t+1>6t+6t\sqrt k$$. But this follows from $$12t^2>12t>12\sqrt k>6\sqrt k$$. (We have $$t>\sqrt k\geq 1$$.)

We are now ready to solve (i). Since $$f_k$$ is increasing, the smallest value of $$\Sigma =f_k(s)$$ is obtained when we take $$s$$ minimal, i.e. when $$s=\sqrt k=2\sin\frac A2$$. Then $$f_k(\sqrt k)=\sqrt{2\sqrt k+2\sqrt k}+\sqrt\frac{4-k}{2\sqrt k+k}$$. But $$\sqrt{2\sqrt k+2\sqrt k}=\sqrt{8\sin\frac A2}$$ and $$\sqrt\frac{4-k}{2\sqrt k+k}=\sqrt\frac{4-k}{\sqrt k(2+\sqrt k)}=\sqrt\frac{2-\sqrt k}{\sqrt k}=\sqrt\frac{2-2\sin\frac A2}{2\sin\frac A2}$$. In conclusion, the smallest possible value of $$\Sigma$$ is $$\sqrt{8\sin\frac A2}+\sqrt\frac{2-2\sin\frac A2}{2\sin\frac A2}=\frac{4\sin\frac A2+\sqrt{2\left( 1-\sin\frac A2\right)}}{\sqrt{2\sin\frac A2}}$$ To see when this minimal value of $$\Sigma$$ is reached, recall that $$k=yz$$ and $$2s=y+z$$ so $$s=\sqrt k$$ happens precisely when $$y=z=\sqrt k$$. This is equivalent to $$zx=xy$$, i.e. $$2\sin^2\frac B2=2\sin^2\frac C2$$, i.e. $$B=C=\frac{\pi-A}2$$. Note that $$A\geq\pi/3$$ implies $$A\geq\frac{\pi-A}2$$ so $$A\geq B=C$$. Hence the condition $$a\geq b\geq c$$ is fulfilled.

For (ii), in order to obtain large values of $$\Sigma =f_k(s)$$ we need large values of $$s$$. Therefore we must find the largest eligible value of $$s$$. Since $$y+z=2s$$, $$yz=k$$ and $$y\leq z$$ we have $$y=s-\sqrt{s^2-k}$$, $$z=s+\sqrt{s^2-k}$$. Since also $$x=\frac{4-k}{2s+k}$$, the condition that $$x\leq y$$ writes as $$\frac{4-k}{2s+k}\leq s-\sqrt{s^2-k}$$, which is equivalent to $$(2s+k)(s-\sqrt{s^2-k})\geq 4-k$$, i.e. $$g_k(s)\geq 4-k$$, where $$g_k:[\sqrt k,\infty )\to{\mathbb R}$$, $$g_k(t)=(2t+k)(t-\sqrt{t^2-k})$$. Moreover, we have $$x=y$$ if and only if $$g_k(s)=4-k$$.

We claim that $$g_k$$ is strictly decreasing. We have $$g_k'(t)=2(t-\sqrt{t^2-k})+(2t+k)(1-\frac t{\sqrt{t^2-k}})=(t-\sqrt{t^2-k})(2-\frac{2t+k}{\sqrt{t^2-k}})$$, which is $$<0$$ for $$t>\sqrt k$$ since the first factor of the product is always positive and the second is always negative. Hence $$g_k$$ is decreasing on its domain $$[\sqrt k,\infty )$$. Also note that $$g_k(\sqrt k)=(2\sqrt k+k)\sqrt k=2k+k\sqrt k\geq 3k\geq 4-k$$, as $$k\geq 1$$. We also have $$\lim_{t\to\infty}g_k(t)=\lim_{t\to\infty}(2t+k)\frac k{t+\sqrt{t^2-k}}=k$$. Hence $$\lim_{t\to\infty}g_k(t)$$ is $$\geq 4-k$$ or $$<4-k$$ when $$k\geq 2$$ or $$k<2$$, respectively. But $$k=4\sin^2\frac A2$$ so the two cases corespond to $$A\geq\pi/2$$ and $$A<\pi/2$$. We consider the two cases separately.

a. $$\pi/3\leq A<\pi/2$$, i.e. $$k<2$$. Then $$g_k(\sqrt k)\geq 4-k>\lim_{t\to\infty}g_k(t)$$. Since $$g_k$$ is decreasing and continuous, there is a unique $$t_0\in [\sqrt k,\infty )$$ such that $$g_k(t_0)=4-k$$. We have $$g_k(t)\geq 4-k$$ if and only if $$t\in [\sqrt k,t_0]$$. Hence the largest value of $$s\in [\sqrt k,\infty )$$ with $$g_k(s)\geq 4-k$$ is $$s=t_0$$ so the largest value of $$\Sigma$$ is achieved for $$s=t_0$$.

If $$s=t_0$$ then $$g_k(s)=4-k$$, which is equivalent to $$x=y$$. Since $$yz=k$$ we have $$z=\frac ky=\frac kx$$. If we replace $$y=x$$ and $$z=\frac kx$$, the identity $$xy+yz+zx+xyz=4$$ writes as $$x^2+k+k+kx=4$$, i.e. $$x^2+kx+2k-4=0$$. But the roots of $$X^2+kX+2k-4=0$$ are $$-2$$ and $$2-k$$. Since $$x>0$$, we have $$x=2-k$$. It follows that $$y=2-k$$ and $$z=\frac k{2-k}$$ and so $$\Sigma =\sqrt x+\sqrt y+\sqrt z=2\sqrt{2-k}+\sqrt\frac k{2-k}$$. But $$2-k=2-4\sin^2\frac A2=2\cos A$$ so $$k=2(1-\cos A)$$. Thus the largest possible value of $$\Sigma$$ is $$2\sqrt{2-k}+\sqrt\frac k{2-k}=\frac{4\cos A+\sqrt{2(1-\cos A)}}{\sqrt{2\cos A}}.$$ In order to achieve this maximal value we need that $$x=y$$, which is equivalent to $$yz=zx$$, i.e. $$4\sin^2\frac A2=4\sin^2\frac B2$$. So we have $$A=B$$ and $$C=\pi-2 A$$. Since $$\pi/3\leq A<\pi/2$$ we have $$A\geq\pi-2A>0$$. Hence $$A=B\geq C>0$$. Hence the condition $$a\geq b\geq c$$ is fulfilled.

b. $$\pi/2\leq A<\pi$$, i.e. $$k\geq 2$$. Then $$\lim_{t\to\infty}g_k(t)\geq 4-k$$. Since $$g_k$$ is strictly decreasing, we have $$g_k(s)>4-k$$ $$\forall s\in [\sqrt k,\infty )$$ so there are no restrictions on $$s$$. Since obviously $$\lim_{s\to\infty}f_k(s)=\infty$$, the value of $$\Sigma =f_k(s)$$ can be arbitrarily large, i.e. there is no finite upper bound.

Alternatively, for every $$M\geq1$$ we may consider the triangle $$ABC$$, where $$c=1$$, $$b=M$$ and $$A$$ is our given value, $$\pi/2\leq A<\pi$$. Since $$M\geq 1$$ we have $$b\geq c$$ and, since $$A\geq\pi/2$$, we have $$a>b=M$$. Thus the condition that $$a\geq b\geq c$$ is fulfilled. Since $$a>M$$, $$b=M$$, $$c=1$$ we have $$\Sigma >\sqrt\frac{a+b-c}c>\sqrt\frac{M+M-1}1=\sqrt{2M-1}.$$ So if $$M\to \infty$$ then $$\Sigma\to\infty$$ so $$\Sigma$$ can be arbitrarily large.