Parameter values that make function values side lengths of a triangle I have been trying to solve the following problem for more than a week without any success.
Given the function:
 $$f(x)=\frac{x^2+mx+4}{x^2+x+4}$$
Find all possible values of the parameter $m$ such that for any three numbers $a,b,c$ the corresponding values of the function $f(a),f(b)$ and $f(c)$ are sides of a triangle.
 A: If $f(a), f(b), f(c)$ are sides of a triangle if they satisfy the triangle inequalities
$$ f(a) + f(b) > f(c), \quad f(a) + f(c) > f(b), \quad f(b) + f(c) > f(a).$$
If we require that this holds for every $a,b,c$, then by symmetry we only need to prove that, for all $a,b,c$, $f(a) + f(b) > f(c)$. Writing out this inequality, we have
$$ \frac{a^2 + ma + 4}{a^2 + a + 4} + \frac{b^2 + mb + 4}{b^2 + b + 4} > \frac{c^2 + mc + 4}{c^2 + c + 4}$$
If we rewrite
$$ \frac{a^2 + ma + 4}{a^2 + a + 4} = \frac{a^2 + a + 4 + (m-1)a}{a^2 + a + 4} = 1 + \frac{(m-1)a}{a^2 + a + 4},$$
and similarly do the same rearrangement for the terms with $b$ and $c$, then our inequality becomes
$$ 2 + \frac{(m-1)a}{a^2 + a + 4} + \frac{(m-1) b}{b^2 + b + 4} > 1 + \frac{(m-1)c}{c^2 + c + 4}.$$
Rearranging, we obtain
$$1 > (m-1) \left[ G(c) - G(b) - G(a) \right]$$
where we have defined a new function $G(x) = x/(x^2 + x + 4)$. Notice that $G(x) \rightarrow 0$ as $|x| \rightarrow \infty$. Therefore we can find the maximum and minimum of $G(x)$ by calculating where its derivative vanishes. We have
$$ G'(x) = \frac{1}{x^2 + x + 4} - \frac{x(2x + 1)}{x^2 + x + 4} = \frac{4 - x^2}{x^2 + x + 4},$$
which vanishes at $x = \pm 2$. We calculate $G(2) = 1/5$ and $G(-2) = -1/3$; therefore these are the maximum and minimum values of $G$, respectively. Therefore we have
$$ - \frac{11}{15} = - \frac{1}{3} - \frac{1}{5} - \frac{1}{5} \le G(c) - G(b) - G(a) \le \frac{1}{5} + \frac{1}{3} + \frac{1}{3} = \frac{13}{15}.$$
Furthermore, these upper and lower bounds are achieved by particular choices of $a,b,c$; therefore, in order for the triangle inequality to be satisfied, therefore, if $m > 1$, we have $(m-1)^{-1} > 13/15$, and if $m < 1$, we have $(m-1)^{-1} > -11/15$. As $m = 1$ is also a solution, then we obtain the set of all possible values of $m$:
$$ -\frac{4}{11} < m < \frac{28}{13}$$
A: Assume that $d$,$e$,$f$ are the lenghts of the sides of a triangle. Drawing sketches of a few different triangles shows that as the angle between $d$ and $e$ approaches 180 degree, $f$ approaches $d+e$ from below, and that as the angle between $d$ and $e$ approaches 0 degree, $f$ approaches $|d-e|$ from ablove. You thus get that $$
  |d-e| < f < d+e
$$
exactly if $d,e,f$ are the lenghts of the sides of a triangle.
Now lets look for intervals $I = (l, u) \subset \mathbb{R}$ where every combination $a,b,c \in I$ fullfills that inequality. Quite obviously, zero cannot be in $I$, since for $f=0$, there are no $d,e$ which fulfill the inequality. More generally, to ensure that $|d-e| < f$ always holds, you must have $l \geq u-l$, since $u-l$ is an upper bound for $|d-e|$ if $d,e \in I$. Similarly, $2l$ is a lower bound for $d+e$ if $d,e \in I$, and thus $u \leq 2l$ must hold if $f < d+e$ is to hold for all $d,e,f \in I$. Since $l \geq u-l$ is in fact equivalent to $2l \geq u$, you get that all $d,e,f \in [l,u]$ are the lenghts of the sides of some triangle iff $$
\begin{eqnarray}
  u &\leq& 2l
\end{eqnarray}
$$
Thus, you need to values of $m$ for which the range of $f$, i.e. the set $\{f(x):x\in\mathbb{R}\}$, is a subset of some inteval which satisfies the above. Can you do that?
