Question

If $a,b$ and $c$ are sides of a triangle, then $\frac{{[a+b+c]}^2}{[ab+bc+ca]}$always belongs to

• A. $[1,2]$
• B. $[2,3]$
• C. $[3,4]$
• D. $[1,3]$

Answer 1

The correct option is C

$[3,4]$

Explanation for the correct option:

Step 1. Using $A.M.\ge G.M.$, we get

$\frac{a^2+b^2}{2}\ge ab,\frac{b^2+c^2}{2}\ge bc$and $\frac{c^2+a^2}{2}\ge ca$

$\Rightarrow$$a^2+b^2\ge 2ab,b^2+c^2\ge 2bc$and $c^2+a^2\ge 2ca$

$\Rightarrow$ $a^2+b^2+c^2\ge ab+bc+\hspace{0.17em}ca$

$\Rightarrow$ ${\left(a+b+\hspace{0.17em}c\right)}^2\ge 3\left(ab+bc+ca\right)$

$\Rightarrow$$\frac{{\left(a+b+\hspace{0.17em}c\right)}^2}{ab+bc+ca}\ge 3$

Step 2. Since $a,b$ and $c$ are sides of a triangle, we get

$\left|a-b\right|\le c,\left|b-c\right|\le a$ and $\left|c-a\right|\le b$

$\Rightarrow$$a^2+b^2-2ab\le c^2,b^2+c^2-2bc\le a^2$ and $c^2+\hspace{0.17em}{a}^2-2ca\le b^2$

$\Rightarrow$ $a^2+b^2+c^2\le 2\left(ab+bc+ca\right)$

$\Rightarrow$ ${\left(a+b+\hspace{0.17em}c\right)}^2\le 4\left(ab+bc+ca\right)$

$\Rightarrow$$\frac{{\left(a+b+\hspace{0.17em}c\right)}^2}{ab+bc+ca}\le 4$

$\mathbf{\therefore }\frac{{\left(a+b+\hspace{0.17em}c\right)}^{\mathbf{2}}}{\mathbf{a}\mathbf{b}\mathbf{}\mathbf{+}\mathbf{}\mathbf{b}\mathbf{c}\mathbf{}\mathbf{+}\mathbf{}\mathbf{c}\mathbf{a}}\mathbf{}\mathbf{\in }\mathbf{}\left[3,4\right]$

Hence, option ‘C’ is correct.