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Let $(a,b,c)$ be the sides of a non-degenerate triangle and $(A,B,C)$ be the angles opposite to these sides respectively. I tried to unify the triangle inequality for sides with the angles and obtained the formulation

$$ \frac{a}{A} + \frac{b}{B} > \frac{c}{C}. $$

Experimental data supports this inequality. Can this be proved or disproved? Also if true, is this inequality known. Looking at its simplicity I presume it must be known. Is there any reference in literature?

Update 11-Nov-2024 : It seems a generalization is possible which I will post as a separate post.

Let $a,b,c$ be the sides of a triangle in any random order, $A,B,C$ be the opposite to these sides respectively, $r$ be the inradius and $R$ be the circumradius; then for all $x \ge 0$ $$ \left(\frac{a}{A}\right)^{\left(\frac{r}{R}\right)^x} + \left(\frac{b}{B}\right)^{\left(\frac{r}{R}\right)^x} \ge \left(\frac{c}{C}\right)^{\left(\frac{r}{R}\right)^x} $$

Nilotpal Sinha
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  • Are values of angles in radians? It will obviously work if $c$ is the biggest side. – Vasili Sep 06 '24 at 13:22
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    @Vasili The inequality is homogeneous, so the units in which the angles are measured do not matter. – jjagmath Sep 06 '24 at 13:24
  • @Vasili In my simulations I used radians but the unit is irrelevant since conversion from one unit to another is basically multiplying all terms by a common factor. – Nilotpal Sinha Sep 06 '24 at 13:26
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    Law of sines is $\frac a{\sin A} = \frac b{\sin B} = \frac c{\sin C}$. As there is no linear conversion between $M$ and $\sin M$ I don't think any such expression is possible. Maybe. – fleablood Sep 06 '24 at 14:52

2 Answers2

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Using law of sines we have

$$\frac{a}{A} + \frac{b}{B} > \frac{c}{C} \iff \frac{\sin A}{A} + \frac{\sin B}{B} > \frac{\sin C}{C}$$

and since for $x\in (0,\pi)$

$$1-\frac x \pi<\frac{\sin x}x<1 \tag 1$$ we have

$$ \frac{\sin A}{A} + \frac{\sin B}{B}>2-\frac A \pi-\frac B \pi=1+\frac C \pi>\frac{\sin C}C$$

To prove $(1)$ for the LHS inequality consider the equivalent

$$f(x)=\sin x-x+\frac{x^2}\pi>0, \; x\in(0,\pi)$$

and by $x=\frac \pi 2+t$ the equivalent

$$g(t)=\cos t+\frac{t^2}\pi-\frac \pi 4>0,\; t\in(0,\pi/2)$$

which is true since $g(0)>0$, $g(\pi/2)=0$ and $g'(t)=\frac{2t}\pi-\sin t<0$ indeed $g'(0)=0$, $g'(\pi/2)=0$ and $g''(t)=\frac 2 \pi -\cos t$ has exactly one root $t_0$ in the interval $(0,\pi/2)$ with $g'(t_0)<0$.

For the RHS inequality recall that $\sin x<x$.

Note that the key inequality, LHS of $(1)$, can also be proved without calculus using references given here for the inequality

$$\sin x >x-\frac{x^3}6, \; x\in(0,\pi/2)\tag 2$$

indeed by symmetry it suffices to prove

$$\sin x-x+\frac{x^2}\pi>0, \; x\in(0,\pi/2)$$

and by $(2)$ we have

$$\sin x-x+\frac{x^2}\pi>x^2\left(\frac{1}\pi-\frac{x}6\right)>0$$

user
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  • The inequality looks very simple that we expect it be there in classical geometry. But it seems the proof requires calculus probably the mean value theorem making it inaccessible to geometers of Euclid's days. I wonder if there is a proof using tools available to Euclid. Also, I will accept your answer. Can you add a reference to the intermediate inequality used to make the proof complete in itself as it is not immediately obvious. – Nilotpal Sinha Sep 07 '24 at 08:57
  • @NilotpalSinha Yes the intermediate inequality requires indeed basic calculus to be proved (I’ll ad that later). I can’t exclude there is a simpler way to obtain the result. – user Sep 07 '24 at 09:53
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    @NilotpalSinha I've added a way which allows to prove the inequality without using calculus. – user Sep 08 '24 at 11:05
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Note that by normalizing, we can set $c=1$. By sine law, we have $a=\dfrac{\sin A}{\sin C}$. Therefore, rearranging we have the statement is equivalent to $$\dfrac{\sin A}{A}+\dfrac{\sin B}{B}-\dfrac{\sin C}{C}>0$$ By angle sum of triangle, we have $C=\pi-A-B$, so this becomes $$\underbrace{\dfrac{\sin A}{A}+\dfrac{\sin B}{B}-\dfrac{\sin(\pi-A-B)}{\pi-A-B}}_{f}>0$$ If we define the above as a function $f(A,B)$, we can see (in geogebra), $f$ is strictly positive in the region $\{(A,B):0<A<\pi,0<B<\pi-A\}$. However, I do not have an analytic way of showing this.

https://www.geogebra.org/3d/cszxb9mj

Angae MT
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