SolFeb2605

If $a$ , $b$ , and $c$ are the sides of a nondegenerate triangle then

${{a}\over{c+b}}+{{b}\over{c+a}}+{{c}\over{b+a}}\leq 2$

Solution I by Bump

Without loss of generality we can take: $a\le b \le c\,$ .

${{a}\over{c+b}}+{{b}\over{c+a}}+{{c}\over{b+a}}\leq {{a}\over{a+b}}+{{b}\over{a+b}}+{{c}\over{a+b}} =1+{{c}\over{a+b}}\le 2$

Since $a+b\ge c\,$ by the triangle inequality.