Let a,b and c be the three sides of a triangle. Show that $\dfrac 1a+\dfrac 1b+\dfrac 1c\le \dfrac 1{a+b-c}+\dfrac 1{a-b+c}+\dfrac 1{-a+b+c}.$

Question

Let a,b and c be the three sides of a triangle. Show that $\dfrac 1a+\dfrac 1b+\dfrac 1c\le \dfrac 1{a+b-c}+\dfrac 1{a-b+c}+\dfrac 1{-a+b+c}.$ When does equality occur?

I have used the Ravi-transformation and obtained

$\frac{1}{x+y} + \frac{1}{x+z} + \frac{1}{y+z} \leq \frac{1}{(x+y)+(y+z)-(x+z)} + \frac{1}{(x+y)-(y+z)+(x+z)} \\ + \frac{1}{-(x+y)+(y+z)+(x+z)} \Longleftrightarrow \frac{1}{x+y} + \frac{1}{x+z} + \frac{1}{y+z} \leq \frac{1}{2x} + \frac{1}{2y} + \frac{1}{2z}.$

But I get stuck here. Does anyone have an idea to proceed?

Answer 1

You are almost there. For $x, y >0$, observe that $$(x-y)^2 \geqslant 0\implies (x+y)^2\geqslant 4xy\implies x+y\geqslant \frac{4xy}{x+y}.$$ Reciprocating yields $$\frac{1}{x+y}\leqslant \frac{x+y}{4xy}\implies \frac{1}{x} + \frac{1}{y} \geqslant \frac{4}{x+y}.$$ Analogous inequalities for the other variables also hold. Summing them up solves the problem.

Answer 2

If I understand correctly you have solved the problem already!

Take: $$ x = \frac{a+b-c}{2} $$ $$ y = \frac{a-b+c}{2} $$ $$ z = \frac{-a+b+c}{2} $$ And note that $x+y = a$ , $x+z = b$ , and $y+z = c$.

The last inequality that you reached is what the problem is asking to prove.