Question

If $a,b,c$ be the sides of a triangle and $P=\frac{{(a+b+c)}^2}{ab+bc+ca}$ then $P\in [1,2]$

• A. True
• B. False

Answer 1

The correct option is B False

$a,b,c$ are sides of triangle.
$\therefore a+b>c,b+c>a,c+a>b$
$\therefore a>|b-c|,b>|c-a|,c>|a-b|$
square and add, we get
$\Rightarrow a^2+b^2+c^2<2(ab+bc+ca)$
$\Rightarrow a^2+b^2+c^2+2(ab+bc+ca)<4(ab+bc+ca)$
$\Rightarrow \frac{{(a+b+c)}^2}{ab+bc+ca}<4$
$\Rightarrow P<4$
Again, ${(a-b)}^2+{(b-c)}^2+{(c-a)}^2\ge 0$
$\Rightarrow \frac{{(a+b+c)}^2}{ab+bc+ca}\ge 3$
$\Rightarrow P\ge 3$
$\therefore 3\le P<4$
or $P\in [3,4)$