Prove that $\frac{ a^2 +b^2 +c^2}{d^2}$ is always greater than $\frac{1}{3}$

Question

Prove that if $a,b,c$ and $d$ are the sides of a quadrilateral ,then the value of

$\frac{ a^2 +b^2 +c^2}{d^2}$ is always greater than $\frac{1}{3}$

Could someone please give me hint to solve this problem.

Answer 1

We have $a+b+c\geq d$ and $$\frac{a+b+c}{3}\leq \sqrt{\frac{a^2+b^2+c^2}{3}}$$Stitch these two inequalities together, and you're done.

Answer 2

As in the comment, $d < a + b +c$. From this: $$ \frac{a^2+b^2+c^2}{d^2}>\frac{a^2+b^2+c^2}{(a+b+c)^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2 +2 a b+2 ac+2bc}\geq \frac{a^2+b^2+c^2}{a^2+b^2+c^2+\left(a^2+c^2\right)+\left(b^2+c^2\right)+(a^2+b^2)}=\frac{1}{3} $$