How can I prove this inequality in a triangle: $\frac{1}{(b+c-a)} +\frac{1}{(c+a-b)} +\frac{1}{(a+b-c)} \gt \frac{1}{a} +\frac{1}{b} +\frac{1}{c}$?

Question

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If $a,b,c$ are the sides of a triangle then show that- $\frac{1}{(b+c-a)} +\frac{1}{(c+a-b)} +\frac{1}{(a+b-c)} \gt \frac{1}{a} +\frac{1}{b} +\frac{1}{c}$

I got this problem from an old book on algebra. I've been trying for a long time but can't reach the answer.

My try on the question-

We know that in a triangle, the sum of two sides is always greater than the third side

$b+c\gt a$ $\implies$ $b+c-a \gt 0$

$c+a\gt b$ $\implies$ $c+a-b\gt 0$

$a+b \gt c$ $\implies$ $a+b-c\gt 0$

Also,

$\frac {1}{(b+c)} \lt \frac{1}{a}$

$\frac{1}{(c+a)} \lt \frac{1}{b}$

$\frac{1}{(a+b)} \lt \frac{1}{c}$

Adding,

$\frac{1}{(b+c)} + \frac{1}{(c+a)} +\frac{1}{(a+b)} \lt \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$

How do I proceed further? All genuine answers are welcome :)

Answer 1

hint: use $\dfrac{1}{x}+\dfrac{1}{y} \ge \dfrac{4}{x+y}$ $ $ $3$ times. The first time for the first two terms with $x = a+b-c, y = b+c-a$.

Answer 2

Your inequality is wrong! Try $a=b=c$.

It should be $$\frac{1}{b+c-a} +\frac{1}{c+a-b} +\frac{1}{a+b-c} \geq \frac{1}{a} +\frac{1}{b} +\frac{1}{c}$$ For the proof we can use the following way.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Since our inequality is fifth degree, we see that $f(w^3)\geq0,$ where $f$ is a linear function,

which says that it's enough to prove our inequality for an extreme value of $w^3$,

which happens in the following cases.

1. $b=c=1$.

We obtain: $$(a-1)^2\geq0;$$

2. $c\rightarrow0^+$ and $b=1$.

We obtain: $$(a+1)(a-1)^2\geq0;$$

3. $a+b-c\rightarrow0^+$

In this case the inequality is obvious.

Done!