Question

If $a,b,c$ 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. $[4,5]$

Answer 1

The correct option is C $[3,4)$

Since $a,b,c>0$ as they are sides of a $△$

We have

$a^2+b^2+c^2-ab-bc-ac=\frac{1}{2}[{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]\ge 0$

$\therefore a^2+b^2+c^2\ge ab+bc+ca$

$\Rightarrow \frac{a^2+b^2+c^2}{ab+bc+ca}\ge \mathrm{1...}\left(1\right)$

Alos $b^2+c^2-a^2=2bc\cos A<2bc$

Similarly $c^2+a^2-b^2<2ac$

Adding these three inequations, ,we get

$a^2+b^2+c^2<2(bc+ac+ab).....\left(2\right)$

from (1) and (2) we have

$1\le \frac{a^2+b^2+c^2}{ab+bc+ca}<2$

$\Rightarrow 1\le \frac{{(a+b+c)}^2}{ab+bc+ca}-2<2$

$\Rightarrow 3\le \frac{{(a+b+c)}^2}{ab+bc+ca}<4$