Question

If $a,b,c,d$ are the sides of a quadrilateral, then the minimum value of $\frac{a^2+b^2+c^2}{d^2}$ is

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

Answer: (C)

$\frac{1}{3}$

$\because {(a-b)}^2+{(b-c)}^2+{(c-a)}^2\ge 0$
$\Rightarrow 2(a^2+b^2+c^2)\ge 2ab+2bc+2ca$
Add $a^2+b^2+c^2$ both sides, we get
$\Rightarrow 2(a^2+b^2+c^2)+(a^2+b^2+c^2)\ge 2ab+2bc+2ca+(a^2+b^2+c^2)$
$\Rightarrow 3(a^2+b^2+c^2)\ge {(a+b+c)}^2>d^2$
$\Rightarrow 3(a^2+b^2+c^2)>d^2$
$\Rightarrow \frac{a^2+b^2+c^2}{d^2}>\frac{1}{3}$
$\therefore$ Minimum value of $\frac{a^2+b^2+c^2}{d^2}$ is $\frac{1}{3}$