Question

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

• A. $\left(\frac{1}{3},\infty \right)$
• B. $\left(\frac{4}{3},\infty \right)$
• C. $(1,\infty )$
• D. $\left(\frac{1}{2},\infty \right)$

Answer 1

The correct option is B $\left(\frac{4}{3},\infty \right)$

Since, $\sum (a-b{)}^2\ge 0$
$\Rightarrow 2(a^2+b^2+c^2)\ge 2(ab+bc+ca)$
Adding $a^2+b^2+c^2$ to both sides
$3(a^2+b^2+c^2)\ge (a+b+c{)}^2>d^2$
$\therefore \frac{a^2+b^2+c^2}{d^2}>\frac{1}{3}$
$\therefore \frac{a^2+b^2+c^2+d^2}{d^2}>\frac{4}{3}$
So, the range of $\frac{a^2+b^2+c^2+d^2}{d^2}$ is $(\frac{4}{3},\infty )$