Question

If $a,b,c$ are the lengths of the sides of a non-equilateral triangle, then, $\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}$ is:

• A. $>\frac{9}{a+b+c}$
• B. $>\frac{1}{a+b+c}$
• C. Both (A) and (B)
• D. None of these

Answer 1

The correct option is C

Both (A) and (B)

Explanation for the correct option:

Finding the value of $\frac{\mathbf{1}}{\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{-}\mathbf{c}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{b}\mathbf{+}\mathbf{c}\mathbf{-}\mathbf{a}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{c}\mathbf{+}\mathbf{a}\mathbf{-}\mathbf{b}}$ using triangle inequality property.

It is given that $a,b,c$ are lengths of a triangle.

$\Rightarrow a>0,b>0,c>0$

By triangle inequality property,

$a+b>c,b+c>a,c+a>b$

$\Rightarrow a+b-c>0,b+c-a>0,c+a-b>0$

Let $x=a+b-c,y=b+c-a,z=c+a-b$

If $m<0$ or $m>1$, then $\left(\frac{{a_1}^m+{a_2}^m+...+{a_n}^m}{n}\right)>{\left(\frac{a_1+a_2+...+a_n}{n}\right)}^m$

$$\begin{gathered} \Rightarrow \frac{x^{-1}+y^{-1}+z^{-1}}{3}>{\left(\frac{x+y+z}{3}\right)}^{-1} \\ \Rightarrow \frac{{\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}+{\displaystyle \frac{1}{z}}}{3}>\frac{3}{x+y+z} \\ \Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}>\frac{9}{x+y+z} \end{gathered}$$

By substituting the values of $x,y,z$, we get

$$\begin{gathered} \frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}>\frac{9}{a+b-c+b+c-a+c+a-b} \\ \Rightarrow \frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}>\frac{9}{a+b+c} \end{gathered}$$

Finding the value of $\frac{\mathbf{1}}{\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{-}\mathbf{c}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{b}\mathbf{+}\mathbf{c}\mathbf{-}\mathbf{a}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{c}\mathbf{+}\mathbf{a}\mathbf{-}\mathbf{b}}$ :

$\begin{array}{rcl}\frac{1}{a+b-c}+\frac{1}{b+c-a}& =& \frac{1}{b-c+a}+\frac{1}{b+c-a}\\ & =& \frac{1}{b-\left(c-a\right)}+\frac{1}{b+\left(c-a\right)}\\ & =& \frac{b-c+a+b+c-a}{b^2-{\left(c-a\right)}^2}\left[by\left(a+b\right)\left(a-b\right)=a^2-b^2\right]\\ & =& \frac{2b}{b^2-{\left(c-a\right)}^2}\end{array}$

As $b^2-{\left(c-a\right)}^2<b^2\Rightarrow \frac{1}{b^2-{\left(c-a\right)}^2}>\frac{1}{b^2}$

$$\begin{gathered} \Rightarrow \frac{2b}{b^2-{\left(c-a\right)}^2}>\frac{2b}{b^2} \\ \Rightarrow \frac{2b}{b^2-{\left(c-a\right)}^2}>\frac{2}{b} \\ \Rightarrow \begin{array}{rcl}& & \frac{1}{a+b-c}\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & \frac{1}{b+c-a}\end{array}\begin{array}{rcl}& >& \end{array}\begin{array}{rcl}& & \frac{2}{b}\end{array}...\left(1\right) \end{gathered}$$

Similarly,

$\frac{1}{b+c-a}+\frac{1}{c+a-b}>\frac{2}{c}...\left(2\right)$

$\frac{1}{a+b-c}+\frac{1}{c+a-b}>\frac{2}{a}...\left(3\right)$

By adding $\left(1\right),\left(2\right)\mathrm{and}\left(3\right)$, we get

$$\begin{gathered} \begin{array}{l}\frac{1}{a+b-c}\end{array}+\begin{array}{l}\frac{1}{b+c-a}\end{array}+\frac{1}{b+c-a}+\frac{1}{c+a-b}+\frac{1}{a+b-c}+\frac{1}{c+a-b}>\frac{2}{a}+\frac{2}{b}+\frac{2}{c} \\ \Rightarrow \begin{array}{l}\frac{2}{a+b-c}\end{array}\begin{array}{l}+\end{array}\begin{array}{rcl}\frac{2}{b+c-a}+\frac{2}{c+a-b}& >& \end{array}\frac{2}{a}+\frac{2}{b}+\frac{2}{c} \\ \Rightarrow \begin{array}{l}\frac{1}{a+b-c}\end{array}\begin{array}{l}+\end{array}\begin{array}{rcl}\frac{1}{b+c-a}+\frac{1}{c+a-b}& >& \end{array}\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \end{gathered}$$

Hence, the correct answer is option C.