Proving an extension to nesbitts inequality: $\frac a{b+c}+\frac b{c+a}+\frac c{a+b}\lt 2$

Question

Prove that $$\frac a{b+c}+\frac b{c+a}+\frac c{a+b}\lt 2$$ given that $a,b,c$ are sides of a triangle.

I know that the above is $\ge \frac32$ but how will you prove the above?

I know this might sound a bit basic but please help.

Answer 1

Hint: Use the substitution $a = x+y, b = y+z, c = z+x$.

Hint: $a + b > x+y+z, b+c > x+y+z, c+a > x+y+z$.

Answer 2

Building on an additional answer, you need to prove that in the process of making substitutions $a = x + y$, $b = y + z$, $c = z + x$ that you can do so using strictly positive $x,y,z$. This will require using properties of a triangle, because in general solving the above system $x,y,z$ for general $a,b,c$ (not sides of a triangle) can give either $x,y$ or $z$ negative. If you can ensure that $x,y,z$ are all positive then the rest should be trivial as indicated by the earlier hints. As a further hint, here are the inverse substitutions that give you $x,y,z$:

$$x = (a-b+c)/2$$ $$y = (a+b-c)/2$$ $$z = (-a+b+c)/2$$