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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 :)

2 Answers 2

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

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    how did you come up with this inequality? – 2017-01-15
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    Use $(x-y)^2\geqslant0$ to prove it – 2017-01-15
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    @DeepSea This inequality is true when $xy\geq0$. Is this condition holds for $a$, $b$ and $c$? – 2017-01-15
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    Yes it is, since $a+b - c > 0$ as these are the lengths of the sides of a triangle . – 2017-01-15
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    Yep this inequality does it...but I still can't understand where that inequality came from – 2017-01-15
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    There are two ways to look at it. First is called the Angel form of Cauchy-Schwarz and second is the variant of the AM-GM . – 2017-01-15
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    I do know the Cauchy Schwarz inequality...are $x,y$ real, positive numbers? – 2017-01-15
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    $(x-y)^2\geq0$ so $x^2+y^2+2xy\geq4xy$ then $\dfrac{x+y}{xy}\geqslant\dfrac{4}{x+y}$ – 2017-01-15
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    alrighty @MyGlasses got it thanx – 2017-01-15
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    @YourAverageEuler "how did you come up with this inequality?" This compares the harmonic mean $h$, defined as $\frac1h=\frac12\left(\frac1x+\frac1y\right)$, to its arithmetic mean $a$, defined as $a=\frac12(x+y)$. For some explanations about the comparisons of $h$, $a$, and the geometric mean $g=\sqrt{xy}$, see [here](https://en.wikipedia.org/wiki/Harmonic_mean#Two_numbers). – 2017-01-15
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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: $$(a-1)^2\geq0;$$

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

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

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

In this case the inequality is obvious.

Done!