Triangle Inequality: Does $a+b \ge C \exp(...)$ Always Hold?

Let's dive into the fascinating world of triangle inequalities! In this exploration, we'll investigate whether the inequality $a+b \ge c \exp\left(\frac{2r}{4R-3r}\right)$ holds true for all triangles, where $a, b, c$ are the sides of the triangle, $R$ is its circumradius, and $r$ is its inradius. This is a captivating question that blends geometry, algebra, and calculus, inviting us to unravel the relationships between different triangle properties.

Background and Motivation

Before we get into the nitty-gritty, it’s helpful to understand the context. Triangle inequalities are fundamental in geometry, offering constraints on the possible side lengths of a triangle. The most basic triangle inequality states that for any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side. This can be expressed as:

$a + b > c$, $b + c > a$, $c + a > b$.

However, mathematicians often explore more intricate relationships involving other triangle properties like the circumradius ($R$) and inradius ($r$). These radii are related to the triangle's area and side lengths, offering a deeper understanding of its geometry. The given inequality,

a+b \ge c \exp\left(\frac{2r}{4R-3r}\\right),

adds an exponential factor involving $R$ and $r$ to the basic triangle inequality, making it a more complex and interesting problem to tackle. It builds upon previously established inequalities, such as $a+b \ge c \exp\left(\frac{r}{2R}\right)$, pushing the boundaries of our understanding.

Exploring the Inequality

Initial Observations

To kick things off, let's consider some initial observations and known results that might help us approach this inequality. First, recall Euler's inequality, which states that $R \ge 2r$ for any triangle. This inequality provides a fundamental relationship between the circumradius and inradius, and it's likely to play a crucial role in our analysis. We can rewrite the exponent in our inequality as follows:

$\frac{2r}{4R-3r}$

Notice that since $R \ge 2r$, we have $4R - 3r > 8r - 3r = 5r > 0$. Thus, the exponent is always positive. This is a good start because it tells us that the exponential term is always greater than 1, meaning that $a+b$ must be greater than $c$ times a factor greater than 1.

Key Relationships and Formulas

To proceed further, let's list some key relationships and formulas that might be useful:

1. Area of a triangle: The area ($A$) can be expressed in multiple ways:
   • $A = \frac{1}{2}ab\sin{C}$,
   • $A = rs$, where $s$ is the semi-perimeter (i.e., $s = \frac{a+b+c}{2}$),
   • $A = \frac{abc}{4R}$.
2. Sine Rule: $\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}} = 2R$.
3. Cosine Rule: $c^2 = a^2 + b^2 - 2ab\cos{C}$.
4. Gerretsen's Inequality: $16Rr - 5r^2 \le s^2 \le 4R^2 + 4Rr + 3r^2$.

These formulas connect the sides, angles, inradius, and circumradius of a triangle, providing tools to manipulate and relate these quantities.

Attempting a Proof

Our goal is to prove that $a+b \ge c \exp\left(\frac{2r}{4R-3r}\right)$. Let's rewrite this as:

$\frac{a+b}{c} \ge \exp\left(\frac{2r}{4R-3r}\right)$.

Using the sine rule, we can express the left side in terms of angles:

$\frac{\sin{A} + \sin{B}}{\sin{C}} \ge \exp\left(\frac{2r}{4R-3r}\right)$.

Now, let's use the sine sum-to-product formula:

$2 \sin{\frac{A+B}{2}} \cos{\frac{A-B}{2}} = 2 \cos{\frac{C}{2}} \cos{\frac{A-B}{2}}$.

So, we have:

$\frac{2 \cos{\frac{C}{2}} \cos{\frac{A-B}{2}}}{\sin{C}} \ge \exp\left(\frac{2r}{4R-3r}\right)$.

Since $\sin{C} = 2 \sin{\frac{C}{2}} \cos{\frac{C}{2}}$, we get:

$\frac{\cos{\frac{A-B}{2}}}{\sin{\frac{C}{2}}} \ge \exp\left(\frac{2r}{4R-3r}\right)$.

This inequality now relates the angles of the triangle to the inradius and circumradius. To proceed, we need to find a connection between $\sin{\frac{C}{2}}$ and $\frac{2r}{4R-3r}$.

Connecting Angles and Radii

We know that $r = 4R \sin{\frac{A}{2}} \sin{\frac{B}{2}} \sin{\frac{C}{2}}$. Therefore,

$\frac{r}{R} = 4 \sin{\frac{A}{2}} \sin{\frac{B}{2}} \sin{\frac{C}{2}}$.

This allows us to express $\sin{\frac{C}{2}}$ in terms of $r$ and $R$, but it also introduces $\sin{\frac{A}{2}}$ and $\sin{\frac{B}{2}}$, making it difficult to isolate the terms we need.

Alternative Approaches

Given the complexity of directly proving the inequality, let's consider alternative approaches. One idea is to examine specific types of triangles, such as equilateral or isosceles triangles, to see if the inequality holds in those cases. If we can find a counterexample, then the inequality is not universally true.

Equilateral Triangles

For an equilateral triangle, $a = b = c$, $A = B = C = 60^\circ$, $R = \frac{a}{\sqrt{3}}$, and $r = \frac{a}{2\sqrt{3}}$. Therefore, $R = 2r$, and our inequality becomes:

$2 \ge \exp\left(\frac{2r}{4(2r)-3r}\right) = \exp\left(\frac{2r}{5r}\right) = \exp\left(\frac{2}{5}\right)$.

Since $e^{2/5} \approx 1.4918$, the inequality holds for equilateral triangles.

Isosceles Triangles

Analyzing isosceles triangles can be more complex, but it may reveal insights into the behavior of the inequality under different conditions. We would need to express the sides and angles in terms of a single variable and then analyze the resulting expression.

Conclusion and Further Research

While we have explored various approaches, a definitive proof or disproof of the inequality $a+b \ge c \exp\left(\frac{2r}{4R-3r}\right)$ remains elusive. The problem combines elements of geometry, trigonometry, and inequality manipulation, making it a challenging yet intriguing problem. Further research might involve exploring more advanced techniques, such as using computer algebra systems to analyze the inequality numerically or searching for related inequalities that could provide a stepping stone towards a solution. This exploration highlights the richness and complexity of triangle inequalities and the ongoing quest to uncover their hidden relationships.

So, does every triangle satisfy this inequality? The journey to find out continues!