A bound for $\sqrt\frac{b+c-a}a+\sqrt\frac{c+a-b}b+\sqrt\frac{a+b-c}c$ in a triangle

Question

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.

Answer 1

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).

Answer 2

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.

Answer 3

(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.

Answer 4

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.