Functions And Triangles


Functions And Triangles - problem

Solution 1

Assume to the contrary that one of the "triangle inequalities" does not hold, say, assume that $x\ge y+z.\,$ Then $x=y+z+t,\,$ $t\ge 0.\,$ We have

$\displaystyle \begin{align} 0&\gt\frac{x}{y+z}-\frac{y}{z+x}-\frac{z}{x+y}\\ &=1+\frac{t}{y+z}-\frac{y}{2z+y+t}-\frac{z}{2y+z+t}. \end{align}$

Now note that the function $\displaystyle 1+\frac{t}{y+z}-\frac{y}{2z+y+t}-\frac{z}{2y+z+t}\,$ is increasing on $[0,\infty),\,$ implying

$\displaystyle f(t)\ge f(0)=1-\frac{y}{2z+y}-\frac{z}{2y+z}=\frac{3yz}{(2z+y)(2y+z)}\gt 0.\,$

A contradiction.

Solution 2

Let $\displaystyle a=\frac{x}{y+z},\,$ $\displaystyle b=\frac{y}{z+x},\,$ $\displaystyle c=\frac{z}{x+y},\,$ $s=x+y+z.\,$ Then, e.g.,

$\displaystyle a=\frac{x}{y+z}=\frac{\displaystyle \frac{x}{s}}{\displaystyle 1-\frac{x}{s}},$

and, subsequently $\displaystyle \frac{x}{s}=\frac{a}{1+a}.\,$ Similarly, $\displaystyle \frac{y}{s}=\frac{b}{1+b}\,$ and $\displaystyle \frac{z}{s}=\frac{c}{1+c}.\,$ We state the following theorem:

If $a,b,c\gt 0\,$ are the sides of a triangle, $I\,$ an interval that includes $0,\,$ $a,\,$ $b,\,$ $c,\,$ and function $f:\,I\to\mathbb{R}\,$ is concave, increasing and satisfies $f(0)=0,\,$ then the numbers $f(a),f(b),f(c)\,$ also form a triangle.

To prove that, note that, for $\lambda\in [0,1]\,$ and $x\in I,\,$

$f(\lambda x)=f(\lambda x+(1-\lambda)0)\ge \lambda f(x)+(1-\lambda)f(0)=\lambda f(x)$

so that $f(\lambda x)\ge \lambda f(x).\,$ If so,

$\displaystyle f(a)=f\left(\frac{a}{a+b}\cdot (a+b)\right)\ge\frac{a}{a+b}\cdot f(a+b)\,\text{and}\\ \displaystyle f(b)\ge \frac{b}{a+b}\cdot f(a+b)\,$

implying that

$\displaystyle f(a)+f(b)\ge \frac{a}{a+b}\cdot f(a+b)+\frac{b}{a+b}\cdot f(a+b)=f(a+b)\ge f(c).$

Similarly, $f(b)+f(c)\ge f(a)\,$ and $f(c)+f(a)\ge b.\,$

Now, function $\displaystyle f(t)=\frac{t}{1+t}\,$ is concave, increasing and satisfies $f(0)=0.\,$ Thus it follows that the quantities $\displaystyle f(a)=\frac{x}{s},\,$ $\displaystyle f(b)=\frac{y}{s},\,$ $\displaystyle f(c)=\frac{z}{s}\,$ form a triangle, and so do $x,y,z.$

Solution 3

Suppose $x$, $y$, and $z$ cannot form a triangle. Thus, the triangle inequality has to be violated for atleast one of the combinations. Without loss in generality, let $z\gt x+y$.


$\displaystyle \begin{align} \frac{x}{y+z}+\frac{y}{z+x} &\lt \frac{x}{x+2y}+\frac{y}{2x+y} \lt \frac{x}{x+y} + \frac{y}{x+y}\\ & =1 \lt \frac{z}{x+y}. \end{align}$

This implies that $\displaystyle\frac{x}{y+z}$, $\displaystyle\frac{y}{z+x}$, and $\displaystyle\frac{z}{x+y}\,$ violate the triangle inequality and cannot form a triangle.


The problem, originally proposed by Seth Easterwood (Lorian Saceanu) at the mathematical inequalities facebook group, has been posted at the CutTheKnotMath facebook page by Leo Giugiuc, along with his solution (Solution 1). I borrowed Solution 2 by Marian Dinca from his post at the mathematical inequalities group.


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