Does every triangle satisfy $a^c + b^c - c^c < \pi$

Question

Let $(a,b,c)$ be the sides of a triangle and let its circumradius be $1$. Is it true that

$$ a^c + b^c - c^c < \pi $$

My progress: In the special case where at least two of the three sides are equal, I have been able to show that the upper bound is about $3.13861$. If two sides are equal and the third side is $c = x$ then the length of the two equal side are $a = b = \sqrt{2 + \sqrt{4-x^2}}$ each. Maximizing the expression using Wolfram Alpha

$$ 2\left(\sqrt{2 + \sqrt{4-x^2}}\right)^x - x^x $$

gives $3.13861$ as the upper bound in this case. Further more, simulation show that this is also the unconditional maxima but I have not been able to prove it. Since $3.13861$ is very close to $\pi$ so, I expressed the above inequality in terms of $\pi$ to present it in an elegant form but it seems the proximity is just a coincidence unless I am missing something.

Answer 1

The three vertices of any such triangle can be taken to be on the unit circle; without loss of generality, one of them can always be $(1,0)$. Therefore the entire configuration space of triangles can be parametrized by two angles $x$ and $y$, with the other two vertices being $(\cos x,\sin x)$ and $(\cos y,\sin y)$. This is a very reasonable numerical optimization problem, and Mathematica finds the maximum value to be about $3.1386122590888217$ when $x=-1.5106639429716202$ and $y=2.3862607049785898$, which is to say when the three vertices are $(1,0)$, $(0.060096151557482914, -0.9981925929238206)$, and $(-0.728044022440954, 0.6855303796244158)$. (The lengths of the sides of this triangle are $1.8590556694615026$, $1.8590556863316139$, and $1.3710607925562726$, so perhaps the optimal triangle is really isosceles and the accuracy isn't as good as the decimal places indicate.) I strongly suspect that the maximum value is some random transcendal number having nothing to do with $\pi$ or any other constant we've ever seen.

Answer 2

Some thoughts.

The conditions are given by $$0 < a, b, c \le 2, \quad a + b > c, \quad b + c > a, \quad c + a > b,$$ $$R = \frac{abc}{\sqrt{(a + b + c)(a + b - c)(b + c - a)(c + a - b)}} = 1. \tag{1}$$

Since $x \mapsto x^{c/2}$ is concave, we have $$a^c + b^c = (a^2)^{c/2} + (b^2)^{c/2} \le 2 \left(\frac{a^2 + b^2}{2}\right)^{c/2}. \tag{2}$$

From (1), we have $$(abc)^2 = (a + b + c)(a + b - c)(b + c - a)(c + a - b)$$ and thus $$(a^2 + b^2 - c^2)^2 = a^2b^2(4 - c^2) \le \frac{(a^2 + b^2)^2}{4}(4 - c^2)$$ which results in $$a^2 + b^2 \le 4 + 2\sqrt{4 - c^2}. \tag{3}$$

Using (2) and (3), we have $$a^c + b^c - c^c \le 2\Big(2 + \sqrt{4 - c^2}\Big)^{c/2} - c^c.$$

Let $f(c) := 2\Big(2 + \sqrt{4 - c^2}\Big)^{c/2} - c^c$. Numerical experiments show that the maximum of $f(c)$ on $[0, 2]$ is near $3.13861$ when $c$ is near $1.371$. Numerical experiments also show that $f$ is concave on $[0, 2]$. Thus, the maximum occurs at $f'(c) = 0$.