Finding the range of $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$, for $a$, $b$, $c$ the sides of a triangle

Question

If $a$, $b$, and $c$ are the three sides of a triangle, then $$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$$ lies in what interval?

Answer 1

Since the target function $f(a,b,c) = \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$ and the triangle inequalities are homogeneous in $a, b, c$, scaling $(a,b,c)$ by a common factor will not change the value of $f(a,b,c)$ nor breaking the triangle inequalities.

This means we only need to study the case $a + b + c = 1$.

When $a + b + c = 1$, the target function $( a,b,c )$ reduces to

$$f(a,b,c)\quad\rightarrow g(a,b,c)\quad\stackrel{def}{=} \frac{a}{1-a} + \frac{b}{1-b} + \frac{c}{1-c}$$ and the triangle inequalities becomes

$$\begin{cases} 0 \le a \le \frac12\\ 0 \le b \le \frac12\\ 0 \le c \le \frac12 \end{cases} $$ The admissible values of $(a,b,c)$ form a triangle on the plane $a + b + c = 1$ with vertices at $$\left(0,\frac12,\frac12\right),\quad\left(\frac12,0,\frac12\right)\quad\text{ and }\quad\left(\frac12,\frac12,0\right)$$ Since $$\frac{d^2}{dx^2} \frac{x}{(1-x)} = \frac{2}{(1-x)^3} > 0\quad\text{ for }\quad x \in \left[0,\frac12\right]$$ $g(a,b,c)$ is a strictly convex function over the triangle. It cannot attain local maximum at any interior points of the triangle nor its edges. This implies

$$g(a,b,c) \le \max\left\{ g\left(0, \frac12,\frac12 \right), g\left(\frac12, 0, \frac12\right), g\left(\frac12,\frac12, 0 \right) \right\} = 2$$

On the other direction, by Jensen's inequality, we have

$$g(a,b,c) \ge g\left(\frac13,\frac13,\frac13\right) = \frac{3}{2}$$

The possible range of $f(a,b,c)$ is $\left[\frac32,2\right]$ or $\left[\frac32,2\right)$ depends on whether degenerate triangle is allowed or not.

Answer 2

Since $a<b+c, b<c+a,c<a+b$ we have that $$ \frac{1}{b+c}\leq \frac{1}{a}, \frac{1}{c+a}\leq \frac{1}{b}, \leq \frac{1}{a+b}\leq \frac{1}{c}, $$ and $$ \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}\leq \frac{a}{a}+\frac{b}{b}+\frac{c}{c}=3 $$

Answer 3

For a $\triangle ABC,$ we have $a+b>c$ and $b+c>a$ and $c+a>b$

For $\bf{Upper\; bound}$ $$a+b>c\Rightarrow 2(a+b)>a+b+c\Rightarrow \frac{1}{2(a+b)}<\frac{1}{a+b+c}\Rightarrow \frac{c}{a+b}<\frac{2c}{a+b+c}$$

In a Similar way for other two expression, And then adding , We get

$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<\frac{2a}{a+b+c}+\frac{2b}{a+b+c}+\frac{2c}{a+b+c} = 2$$

Similarly For $\bf{Lower\; bound}$

$$a>0\Rightarrow a+b+c>b+c\Rightarrow \frac{1}{a+b+c}<\frac{1}{b+c}\Rightarrow \frac{a}{a+b+c}<\frac{a}{b+c}$$

Similar way for other two expression, And then Adding, We get

$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c} = 1$$

So we get $$1<\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$$

Answer 4

As expression

$$\tag{1}f(a,b,c):=\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$$

is invariant in any change of scale, we introduce the following (value) constraint:

$$\tag{2}g(a,b,c):=a+b+c=1$$

This means that the subsequent work is in this simplex, or more exactly in the subset of this simplex with supplementary (domain) constraints:

$$\tag{3} a \leq b+c, \ \ b \leq c+a, \ \ c \leq a+b$$

(in case of equality, we have flat triangles).

Under all these constraints, $(1)$ becomes:

$$\tag{4}f(a,b,c)=\frac{a}{1-a} + \frac{b}{1-b} + \frac{c}{1-c}$$

maximizing/minimizing $(4)$ under constraint $(2)$ is done by writing Lagrange equations:

$$grad (f) \ = \ \lambda \ grad(g) \ \ \ \iff \ \ \ \cases{\dfrac{1}{(1-a)^2}=\lambda 1\\ \dfrac{1}{(1-b)^2}=\lambda 1\\ \dfrac{1}{(1-c)^2}=\lambda 1},$$

from which $a=b=c=\frac13$ (equilateral triangle), giving $f(\frac13,\frac13,\frac13)=\frac32.$

This extremum is a minimum because the Hessian matrix,

diag $\left(\dfrac{2}{(1-a)^3},\dfrac{2}{(1-b)^3},\dfrac{2}{(1-c)^3}\right)$ is positive definite.

Now, the maxima(s) should be found on the borders. As all the borders play a symmetrical role, consider only boundary $c=a+b$, which implies (using $(2)$)

$$1=a+b+c=2(a+b)\ \ \ \Longrightarrow \ \ \ a+b=\frac12$$

$$\tag{5} \Longrightarrow \ \ \ 0 \leq a\leq 1/2, b=\dfrac12-a, c=\frac12.$$

Thus:

$$f(a,b,c)=f(a,\frac12-a,\frac12)=\dfrac{a}{1-a}+2-2a.$$

A rapid study of this function for $a \in [0,\frac12]$ shows that the value of its maximum is $2$, occurring for $a=0$ or $a=\frac12$. Both associated with a "flat" triangle with sides $(\frac12,\frac12,0) $.

Conclusion: the values taken by $f$ are situated in the interval $[\frac32,2]$. As $f$ is continuous on a compact domain, it is the whole interval $[\frac32,2].$

Answer 5

The problem is Nesbitt's inequality. (Source: A.M. Nesbitt, Problem 15114, Educational Times (2) 3, (1903), 37-38.)

Lower Bound: Let's say $S= \dfrac{a}{b+c} + \dfrac{b}{c+a} + \dfrac{c}{a+b} $. $S+3= \dfrac{a}{b+c}+1 + \dfrac{b}{c+a}+1 + \dfrac{c}{a+b}+1=(a+b+c)\left( \dfrac{1}{b+c} + \dfrac{1}{c+a} + \dfrac{1}{a+b}\right)$. By aritmethic-harmonic mean inequality we can find that $$ (a+b+c)\left( \dfrac{1}{b+c} + \dfrac{1}{c+a} + \dfrac{1}{a+b}\right) \ge \dfrac92$$ Therefore $S\ge \dfrac32$.Equality holds when $a=b=c$.

For other solutions of first part: Nesbitt's Inequality

Upper Bound: 1. Method: By triangle inequality $b+c<a, c+a<b, a+b<c$ and hence $b+c>\dfrac{1}{2}(a+b+c)$, $c+a>\dfrac{1}{2}(a+b+c)$, $a+b>\dfrac{1}{2}(a+b+c)$. We have $$ S = \dfrac{a}{b+c} + \dfrac{b}{c+a}+\dfrac{c}{a+b} < \dfrac{2a}{a+b+c} + \dfrac{2b}{a+b+c} + \dfrac{2c}{a+b+c} = \dfrac{2(a+b+c)}{a+b+c} = 2 $$

If $a=b$ and $c \to 0$ degenere triangle form, then $S=2$. Hence for all non-degenere triangles $S<2$.

2. Method: Without loss of generality, we assume that $a\leq b \leq c$. Thus, $$ \dfrac{a}{b+c}\leq \dfrac{a}{a+c}, \quad \dfrac{b}{a+c}\leq \dfrac{c}{a+c}, \quad c < a+b $$ and therefore $$S < \dfrac{a}{a+c} + \dfrac{c}{a+c} + \dfrac{a+b}{a+b} = \dfrac{a+c}{a+c} + 1 = 2 .$$ So, $S<2$.

In summary, we obtain that $$ \dfrac32 \le S <2 .$$

Also this interval is best.