Question

If $a,b,c$ are the sides of a triangle then the range of $\frac{ab+bc+ca}{a^2+b^2+c^2}$ is

• A. $[1,2)$
• B. $(\frac{1}{2},1]$
• C. $[\frac{1}{2},2)$
• D. $(-\frac{1}{2},2)$

Answer 1

Answer: (B)

$(\frac{1}{2},1]$

For a triangle$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$
Therefore
$a=\frac{c\left(\sin A\right)}{\sin C}$ and $b=\frac{c\left(\sin B\right)}{\sin C}$
By substituting we get the above expression as
$\frac{\sin A\sin B+\sin B\sin C+\sin C\sin A}{{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C}$
Multiplying and dividing by 2 we get
$\frac{2(\sin A\sin B+\sin B\sin C+\sin C\sin A)}{2({\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C)}$
$=\frac{1}{2}\left[\frac{(\sin A+\sin B+\sin C{)}^2-({\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C)}{{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C}\right]$
$=\frac{1}{2}[\frac{(\sin A+\sin B+\sin C{)}^2}{{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C}-1]$
Now we know that for maximum value of triangle must be equilateral
Substituting$\sin A=\sin B=\sin C=\frac{\sqrt{3}}{2}$, we get $\frac{1}{2}[\frac{9}{3}-1]$
$=\frac{1}{2}[3-1]$
$=1$
Therefore maximum value of the above expression is 1.Hence, option 'B' is correct.