<s>[INST] How can I prove this inequality in a triangle: $\frac{1}{(b+c-a)} +\frac{1}{(c+a-b)} +\frac{1}{(a+b-c)} \gt \frac{1}{a} +\frac{1}{b} +\frac{1}{c}$? 
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 :)
 [/INST] 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.


*

*$b=c=1$.


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


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


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


*$a+b-c\rightarrow0^+$
In this case the inequality is obvious.
Done!
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