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.
Best Answer
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.
$\phantom{2}$
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.