Does every triangle satisfy $\frac{abc}{R^3} \ln \left(\frac{a}{R}\right)\ln \left(\frac{b}{R}\right)\ln \left(\frac{c}{R}\right) > -\ln 2$?

Question

If $a,b,c$ are the sides of a triangle inscribed in a circle of radius $R$. Is it true that

$$ \frac{abc}{R^3} \ln \left(\frac{a}{R}\right)\ln\left(\frac{b}{R}\right)\ln \left(\frac{c}{R}\right) > - \ln 2 \tag 1 $$

I ran a Monte Carlo simulation for the expression on the LHS and obtained a minima of $- 0.6923259$ after a billion samples. This value is slightly greater than $\ln2 \approx 0.6931$. Can this proved?

Update 1: Using the insight from the comment made by @Anon we can show the slightly weaker result that LHS $> - \frac{4\ln^2 2}{e} \approx -0.7069951$. If one side of the triangle is less than $R$ and the other two sides are greater than $R$ then the LHS will be negative. Hence we need to minimize $x \ln x$ for one side and maximize it for the other two sides. The minimum value of $x\ln x$ occurs at $x = \frac{R}{e}$ and for the triangle of circum radius $R$, since no side can exceed $2R$ hence $x\ln x < \frac{2R}{R} \ln \frac{2R}{R} = 2\ln 2$. Hence the LHS of $(1)$ must be at least $> - \frac{4\ln^2 2}{e}$.

Update 2: An improvement of the estimate in update 1 is as follows. If one side of a triangle is $a = \frac{1}{e}$, what is the maximum possible length of the other sides. Simple algebraic manipulations give $b = c = \sqrt{2 + \sqrt{4 - \frac{1}{e^2}}}$ and substituting these in $(1)$ we get

$$ LHS \ge -\frac{1}{e}\left(\sqrt{1 - \frac{1}{2e}} + \sqrt{1 + \frac{1}{2e}}\right)^2 \ln^2 \left(\sqrt{1 - \frac{1}{2e}} + \sqrt{1 + \frac{1}{2e}}\right) \approx -0.692325 $$

But must be noted that this may not be the minima since the true minima may occur at $a \ne \frac{1}{e}$.

Update 3: Under the assumption that the minima will occur when the triangle is isosceles, we can prove that the minima is $-0.62918$ as shown by K.defaoite. I have separate and proof which I have posted below.

Answer 1

Proof.

We need to prove that, for all $x, y, z > 0$ with $x + y + z = \pi$, $$8\sin x \sin y \sin z \cdot \ln (2\sin x) \cdot \ln(2 \sin y) \cdot \ln(2\sin z) > -\ln 2. \tag{1}$$

We only need to prove the case that $\ln (2\sin x) \cdot \ln(2 \sin y) \cdot \ln(2\sin z) < 0$.

We split into two cases.

Case 1: $\ln(2\sin x) < 0, \ln(2\sin y) < 0, \ln(2\sin z) < 0$

It is easy to prove that $-\mathrm{e}^{-1} \le u\ln u < 0$ for all $u\in (0, 1)$. Thus, we have $-\mathrm{e}^{-1} \le 2\sin x \ln(2\sin x) < 0$ etc. and $$\mathrm{LHS} \ge (-\mathrm{e}^{-1})^3 > -\ln 2.$$

$\phantom{2}$

Case 2: $\ln(2\sin x) > 0, \ln(2\sin y) > 0, \ln(2 \sin z) < 0$

We have $x, y \in (\pi/6, 5\pi/6)$ and $z \in (0, \pi/6)$.

Note that $x\mapsto \ln (2\sin x)$ is concave on $(\pi/6, 5\pi/6)$. We have $$\ln(2\sin x) + \ln(2\sin y) \le 2\ln\left(2\sin \frac{x + y}{2}\right). \tag{2} $$ Using (2), we have $$2\sin x \cdot 2\sin y \le \left(2\sin \frac{x + y}{2}\right)^2. \tag{3}$$ By AM-GM, using (2), we have $$ \ln(2\sin x) \ln (2\sin y) \le \frac14[\ln(2\sin x) + \ln(2\sin y)]^2\le \ln^2\left(2\sin \frac{x + y}{2}\right). \tag{4}$$

From (1), (3) and (4), it suffices to prove that $$ \left(2\sin \frac{x + y}{2}\right)^2\cdot \ln^2\left(2\sin \frac{x + y}{2}\right) \cdot 2\sin z \cdot \ln(2\sin z) > -\ln 2$$ or $$ \left(2\cos\frac{z}{2}\right)^2\cdot \ln^2\left(2\cos \frac{z}{2}\right) \cdot 2\sin z \cdot \ln(2\sin z) > -\ln 2$$ or $$\ln^2 2 > \left(2\cos\frac{z}{2}\right)^4\cdot \ln^4\left(2\cos \frac{z}{2}\right) \cdot 4\sin^2 z \cdot \ln^2(2\sin z)$$ or $$\ln^2 2 > (2 + 2\cos z)^2 \cdot \frac{1}{2^4}\ln^4 (2 + 2\cos z)\cdot 4(1 - \cos^2 z)\cdot \frac{1}{2^2}\ln^2(4 - 4\cos^2 z)$$ where we use $4\cos^2 \frac{z}{2} = 2 + 2\cos z$.

Letting $u = \cos z$, it suffices to prove that, for all $\frac{\sqrt 3}{2} < u < 1$, $$\ln^2 2 > (2 + 2u)^2 \cdot \frac{1}{2^4}\ln^4 (2 + 2u)\cdot 4(1 - u^2)\cdot \frac{1}{2^2}\ln^2(4 - 4u^2) \tag{5}$$ which is true. The proof is given at the end.

We are done.

$\phantom{2}$

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Proof of (5):

Using $\frac{64}{63}u^2 + \frac{63}{64} - 2u = \frac{1}{4032}(64u-63)^2 \ge 0$, we have $$(2 + 2u)^2 = 4 + 4u^2 + 8u \le 4 + 4u^2 + 4\left(\frac{64}{63}u^2 + \frac{63}{64}\right) .$$

It suffices to prove that $$\ln^2 2 > \frac{1}{256}\left(\frac{508}{63}u^2 + \frac{127}{16}\right)\ln^4 \left(\frac{508}{63}u^2 + \frac{127}{16}\right) \cdot (1 - u^2)\ln^2(4 - 4u^2). \tag{6}$$

Letting $4 - 4u^2 = v$, it suffices to prove that, for all $v \in (0, 1)$, $$\ln^2 2 > \frac{1}{1024}\left(\frac{16129}{1008} - \frac{127}{63}v\right) \ln^4 \left(\frac{16129}{1008} - \frac{127}{63}v\right) \cdot v\ln^2 v. \tag{7}$$

We can prove that, for all $v\in (0, 1)$, $$v\ln^2 v \le \frac98\ln^2 2 + (9\ln^2 2 - 6\ln 2)(v - 1/8). \tag{8}$$ (Note: The RHS is the first order Taylor approximation of LHS around $v = 1/8$. Proof: Take derivative.)

We can also prove that, for all $v\in (0, 1)$, $$\ln \left(\frac{16129}{1008} - \frac{127}{63}v\right) \le \ln\frac{15875}{1008} - \frac{16}{125}(v - 1/8). \tag{9}$$ (Note: The RHS is the first order Taylor approximation of LHS around $v = 1/8$. Proof: Take derivative.)

It suffices to prove that, for all $v\in (0, 1)$, \begin{align*} \ln^2 2 &> \frac{1}{1024}\left(\frac{16129}{1008} - \frac{127}{63}v\right) \left[\ln\frac{15875}{1008} - \frac{16}{125}(v - 1/8)\right]^4\\[10pt] &\qquad \times \left[\frac98\ln^2 2 + (9\ln^2 2 - 6\ln 2)(v - 1/8)\right]. \tag{10} \end{align*}

Using $\ln\frac{15875}{1008} < \frac{397}{144}$, it suffices to prove that for all $v\in (0, 1)$, \begin{align*} \ln^2 2 &> \frac{1}{1024}\left(\frac{16129}{1008} - \frac{127}{63}v\right) \left[\frac{397}{144} - \frac{16}{125}(v - 1/8)\right]^4\\[10pt] &\qquad \times \left[\frac98\ln^2 2 + (9\ln^2 2 - 6\ln 2)(v - 1/8)\right] \end{align*} or \begin{align*} 1 &> \frac{1}{1024}\left(\frac{16129}{1008} - \frac{127}{63}v\right) \left[\frac{397}{144} - \frac{16}{125}(v - 1/8)\right]^4\\[10pt] &\qquad \times \left[\frac98 + \left(9 - \frac{6}{\ln 2}\right)(v - 1/8)\right]. \tag{11} \end{align*}

We split into two cases.

Case 1: $0 < v \le 1/8$

Using $\ln 2 > \frac{70}{101}$, it suffices to prove that \begin{align*} 1 &> \frac{1}{1024}\left(\frac{16129}{1008} - \frac{127}{63}v\right) \left[\frac{397}{144} - \frac{16}{125}(v - 1/8)\right]^4\\[10pt] &\qquad \times \left[\frac98 + \left(9 - \frac{6}{70/101}\right)(v - 1/8)\right].\tag{12} \end{align*} Letting $v = \frac{1}{8 + w}$ for $w \ge 0$, (12) is written as $$\frac{F(w)}{10113169489920000000000000(8+w)^6} > 0$$ where \begin{align*} F(w) &= 2416264063255035327931w^6 + 53231769508888881467040w^5\\ &\qquad + 440259554245249137560000w^4 + 4667308406207884160000000w^3\\ &\qquad + 67053001632350760000000000w^2 + 463246417085920000000000000w\\ &\qquad + 1111827500508480000000000000. \end{align*} Thus, (12) is true.

Case 2: $1/8 < v < 1$

Using $\ln 2 < \frac{61}{88}$, it suffices to prove that \begin{align*} 1 &> \frac{1}{1024}\left(\frac{16129}{1008} - \frac{127}{63}v\right) \left[\frac{397}{144} - \frac{16}{125}(v - 1/8)\right]^4\\[10pt] &\qquad \times \left[\frac98 + \left(9 - \frac{6}{61/88}\right)(v - 1/8)\right]. \tag{13} \end{align*} Letting $v = \frac{1 + 8w}{8 + 8w}$ for $w > 0$, (13) is written as $$\frac{G(w)}{17625809682432000000000000(1+w)^6} > 0$$ where \begin{align*} G(w) &= 823425037465312092776856w^6 + 3524126527531263109253089w^5\\ &\qquad +5778219879301331456143375w^4+4426205206524727153031250w^3\\ &\qquad +1504381676597156324218750w^2+163085374636585205078125w\\ &\qquad +7391954425377685546875. \end{align*} Thus, (13) is true.

We are done.

Answer 2

Partial answer / simplification.

WLOG consider $R=1$. Other values of $R$ follow by simple scaling.

Consider a unit circle just touching the origin, i.e consider the set $\{(x,y)\in\Bbb R^2\mid \Vert(x-1,y)\Vert=1\}$. Again, WLOG, we can consider one of the points to be precisely at the origin $(0,0)$, as all the other cases follow by a simple rotation.

So, we have reduced our four-parameter problem (a,b,c,R) into a two-parameter problem. Next, any triangle in the unit circle can be described by three points. The points are $$P_0=(0,0) \\ P_1=(1+\cos\alpha,\sin\alpha) \\ P_2=(1+\cos\beta,\sin\beta) \\ \text{where}~\alpha,\beta\in(-\pi,\pi) ~\text{and}~\alpha\neq \beta$$ Getting the side lengths is easy. We call $$a=\Vert P_1-P_0\Vert=2\cos(\alpha/2) \\ b=\Vert P_2-P_0\Vert=2\cos(\beta/2) \\ c=\Vert P_2-P_1\Vert=\sqrt{2-2\cos(\alpha-\beta)}$$

Therefore, we can equivalently seek $$\inf_{|\alpha|<\pi \\ |\beta|<\pi}\left\{4\cos(\alpha/2)\cos(\beta/2)\sqrt{2-2\cos(\alpha-\beta)}~\log(2\cos(\alpha/2))\log(2\cos(\beta/2))\log\sqrt{2-2\cos(\alpha-\beta)}\right\} \\ =\inf_{|\alpha|<\pi \\ |\beta|<\pi}\left\{f\big(2\cos(\alpha/2)\big)f\big(2\cos(\beta/2)\big)f\left(\sqrt{2-2\cos(\alpha-\beta)}\right)\right\}$$

Putting this minimization problem into Mathematica using the command

f[z_] := z*Log[z]
NMinimize[{f[2*Cos[\[Alpha]/2]]*f[2*Cos[\[Beta]/2]]*
   f[Sqrt[2 - 2 Cos[\[Alpha] - \[Beta]]]], \[Alpha] \[Element] 
   Reals, \[Beta] \[Element] Reals, Abs[\[Alpha]] < \[Pi], 
  Abs[\[Beta]] < \[Pi]}, {\[Alpha], \[Beta]}]

It returns the value of $-0.692918$ at $\alpha=-\beta=0.177529$, which is just barely bigger than $-\log 2$.

I think if you can prove that the minimum has to occur on the line $\alpha=-\beta$, then you should be able to prove that the minimum is definitely larger than $-\log 2$. Maybe the best way to do that is to let $$F(x,y)=f(2\cos(x/2))f(2\cos(y/2))f(\sqrt{2-2\cos(x-y)})$$ Explicitly calculate $\nabla F$, and show that the only suitable solutions to $\nabla F(x,y)=0$ are on the line $\alpha=-\beta$.

Answer 3

Let $a,b,c$ be the sides of a triangle and $R$ be its circumradius. Let $f\left(\frac{a}{R}, \frac{b}{R}, \frac{c}{R}\right)$ whose maxima or minima we are interested. WLOG we can take $R = 1$ since other values of $R$ follow by scaling. Thus we are interested in the extreme values of $f(a,b,c)$.

Assumption: If $f(a,b,c)$ is symmetric then the extreme values will occur when the triangle is isosceles.

I do not have a proof of this but intuitively this seems plausible since for a symmetric function, interchanging any two sides does not alter the value of $f$. In many cases, the extreme values occur when all three sides are equal but equilateral triangles are a subset of isosceles triangles so it does not violate the above assumption.

Let $a$ be the base of the isosceles triangle which gives the extreme values and let the other two equal sides be of length $x$ each. Since the circumradius $R = 1$ using equation 1 in this link we have

$$ \frac{x^2}{\sqrt{(2x+a)(2x-a)}} = 1 \tag 1 $$

Solving for $x$ we get $x = \sqrt{2 \pm \sqrt{4-a^2}}$ as the feasible solutions. Hence the extreme values of $f(a,b,c)$ will occur at $f\left(a, \sqrt{2 \pm \sqrt{4-a^2}}, \sqrt{2 \pm \sqrt{4-a^2}}\right)$. Since $a$ is the side of a triangle inscribed in circle of radius $R=1$, the extreme values will be the local maxima and minima for $0 \le a \le 2$. Plugging these values of $a,b,c$ in $f(a,b,c) = abc \ln a \ln b \ln c$, the minimum value is the local minima of

$$ (a\ln a) \left(\sqrt{2 \pm \sqrt{4-a^2}}\right)^2 \ln \left(\sqrt{2 \pm \sqrt{4-a^2}}\right) \tag 2 $$

for $a \le 2$ which is $-0.692918$ at $a = 0.353196$ as calculated by Wolfram Alpha.

Additionally solving $(2)$ for maxima gives the maximum value of $0.861242$ at $a = b = c = \sqrt{3}$ i.e. equilateral triangle

Answer 4

Does every triangle satisfy

$$ \frac{abc}{R^3} \ln \frac{a}{R}\ln \frac{b}{R}\ln \frac{c}{R} > - \ln 2 $$

where $R$ is the circumradius?

Substituting $\frac{a}{R}=2\sin\alpha, \frac{b}{R}=2\sin\beta, \frac{c}{R}=2\sin\gamma$ this becomes

$$ 8\sin\alpha\sin\beta\sin\gamma\ln(2\sin\alpha)\ln(2\sin\beta)\ln(2\sin\gamma) > - \ln 2 $$

Replacing $\gamma=\pi-(\alpha+\beta)$ gives

$$ 8\sin\alpha\sin\beta\sin(\alpha+\beta)\ln(2\sin\alpha)\ln(2\sin\beta)\ln(2\sin(\alpha+\beta)) > - \ln 2 $$

So this reduces to finding the minimum value of a function

$$ f(x,y)=8\sin(x)\sin(y)\sin(x+y)\ln(2\sin(x))\ln(2\sin(y))\ln(2\sin(x+y)) $$ under the restraint $0<x+y<\pi$.

Wolfram Alpha computes a minimum value of \begin{eqnarray}f_\min&=&-.6929184\\ &=&\ln(0.5001144034)\\ &=&-\ln(2)+0.000285555 \end{eqnarray}at $(x,y)\approx(0.177529,1.48203)$

So it seems that the answer is yes.