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# If a,b,c are the sides of a triangle, then the statement, “$\large\frac{1}{(b+c)},\: \large\frac{1}{(c+a)},\: \large\frac{1}{(a+b)}$ are also the sides of the triangle”, is

$\begin {array} {1 1} (A)\;tautology & \quad (B)\;fallacy \\ (C)\;cannot \: be\: determined & \quad (D)\;None\: of\: these \end {array}$

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A)
Assume that $a \geq b \geq c$. We must have $b+c > a.$
Also note that $b+c \leq c+a \leq a+b$
$\large\frac{1}{(b+c)} \geq \large\frac{ 1}{(c+a)} \geq \large\frac{1}{(a+b)}$
To show that $\large\frac{1}{(b+c)},\large\frac{1}{(c+a)},\large\frac{1}{(a+b)}$ are sides of a triangle,
$\large\frac{1}{(c+a)} + \large\frac{1}{(a+b)} > \large\frac{1}{(b+c)}$……………(i)
As $a \geq b \geq c$, we get
$2a \geq a+b\: and\: 2a \geq a+c$
$\large\frac{1}{2a} \leq \large\frac{1}{(a+b)}, \large\frac{1}{2a} \leq \large\frac{1}{(a+c)}$
$\large\frac{1}{(a+b)} + \large\frac{1}{(a+c)} \geq \large\frac{1}{2a} + \large\frac{1}{2a} = \large\frac{1}{a} > \large\frac{1}{(b+c)}$ [From Eq. (i)]