Don't Guess, Solve it!

Using Ravi subs $a=y+z,~b=z+x,~c=x+y$ , our expression changes to $\dfrac{y+z}{2y}+\dfrac{z+x}{2z}+\dfrac{x+y}{2x}=\dfrac{1}{2}\left(\dfrac{x}{z}+\dfrac{y}{x}+\dfrac{z}{y}\right)+\dfrac{3}{2}.$ The bracket part is $\ge 3$ by AM-GM, so the whole expression is $\ge \fbox{3.}$

Equality occurs for $x=y=z$ , that is for equilateral triangles.

Sum of any 2 sides is always greater than the third side.

$\frac{a + c - b + b - c}{a + c - b} + \frac{a + b - c + c - a}{a - c + b} + \frac{c + b - a + a - b}{b + c - a}$

$= 3 + \frac{b - c}{a + c - b} + \frac{ c - a}{a - c + b} + \frac{a - b}{b + c - a}$

As said above, we can see that the denominator will remain positive.

• Case 1 - a>b>c , we can see one will be negative for other cases also ( b>c>a) , (c>b>a),(b>a>c),(c>a>b). So the 2 will add up and will always be greater than the third one.

  • Case 2 - a = b = c , 3 + 0 = 3

$\frac{a}{c + a - b} + \frac{b}{a + b - c} + \frac{c}{b + c - a} \geq 3$