# Inequality involving sides of a triangle

How can one show that for triangles of sides $a,b,c$ that

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

My proof is long winded, which is why I am posting the problem here.

Step 1: let $a=x+y$, $b=y+z$, $c=x+z$, and let $x+y+z=1$ to get

$\frac{1-x}{1+x}+\frac{1-y}{1+y}+\frac{1-z}{1+z}<2$

Step 2: consider the function $f(x)=\frac{1-x}{1+x}$, and note that it is convex on the interval (0,1), so the minimum of

$\frac{1-x}{1+x}+\frac{1-y}{1+y}+\frac{1-z}{1+z}$

is reached when the function takes the extreme points. i.e. $x=y=0, z=1$.

But going back to the fact that this is a triangle, we note that $x=y=0 \implies a=0$ which is not possible, so the inequality is strict.

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Note that this inequality is "sharp". With the triangle having sides $a=b$, and $c$ arbitrarily close to 0, you have the left-hand side arbitrarily close to 2. – alex.jordan Jul 30 '11 at 17:52

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

since $a<b+c$ etc.

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Very nice. ${}$ – joriki Jul 30 '11 at 17:48
@Dave Radcliffe, do you need any of $a < b+c, \cdots?$ is not $a, b, c > 0$ enough? does't $\frac{a}{b} < \frac{a+c}{b+c}$ always hold positive $a, b$ and $c?$? – abel Nov 26 '14 at 6:13
For positive $a$, $b$, and $c$, $\frac{a}{b} < \frac{a+c}{b+c} \iff a(b+c) < b(a+c) \iff ac < bc \iff a<b$. – Dave Radcliffe Nov 27 '14 at 1:40
@Dvid Radcliffe, i got it. thanks. – abel Nov 27 '14 at 6:14

Let $c$ be the largest of $a$, $b$, and $c$. Then \begin{align} \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}&\le\frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{a+b}\\ &=1+\frac{c}{a+b}\\ &< 2 \end{align} since $c<a+b$.

For a degenerate triangle with $a=0$, we have $c=a+b$ and the sum equals $2$.

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See the deleted comment (pls do not share it, I wanna add it to my book) :-) – user 1618033 Jan 7 '15 at 14:20