Question

. If $a,b,c$ are the sides of a non-equilateral triangle, then the expression $(b+c–a)(c+a–b)(a+b–c)–abc$ is:

• A. Negative
• B. Non-negative
• C. Positive
• D. Non-positive

Answer 1

The correct option is A

Negative

Explanation for the correct option:

Finding the value of $\mathbf{(}\mathit{b}\mathbf{+}\mathit{c}\mathbf{–}\mathit{a}\mathbf{)}\mathbf{(}\mathit{c}\mathbf{+}\mathit{a}\mathbf{–}\mathit{b}\mathbf{)}\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathbf{–}\mathit{c}\mathbf{)}\mathbf{–}\mathit{a}\mathit{b}\mathit{c}$:

It is given that $a,b,c$ are the sides of a triangle,

$\Rightarrow a,b,c>0$ and by triangle inequality property,

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

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

Let $x=a+b-c,y=b+c-a,\mathrm{and}z=c+a-b$

$\Rightarrow x>0,y>0,z>0$

By $AM-GM$ inequality property, for positive distinct numbers $AM>GM$.

Therefore,

$$\begin{gathered} b=\frac{x+y}{2}>\sqrt{xy} \\ c=\frac{y+z}{2}>\sqrt{yz} \\ a=\frac{z+x}{2}>\sqrt{zx} \end{gathered}$$

$\begin{array}{rcl}& \Rightarrow & abc>\sqrt{xy}\times \sqrt{yz}\times \sqrt{zx}\\ & >& \sqrt{x^2{y}^2{z}^2}\\ & >& xyz\end{array}$

By substituting the values of $x,y,\mathrm{and}z$, we get

$$\begin{gathered} \begin{array}{rcl}abc& >& \left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\end{array} \\ \Rightarrow \begin{array}{l}\left(a+b-c\right)\end{array}\begin{array}{l}\left(b+c-a\right)\end{array}\begin{array}{l}\left(c+a-b\right)\end{array}-abc<0 \end{gathered}$$

Therefore, the correct answer is option A.