Side relationship of a quadrilateral

Question

If $p,q,r,s$ are the sides of quadrilateral. Find the minimum value of $$\frac{p^2 + q^2 + r^2}{s^2}.$$

I used AM GM INEQUALITY but failed.

Please see whether it's right or wrong approach to solve the problem.

Answer 1

By the triangle inequality $\,p+q+r \ge s \iff \dfrac{p+q+r}{s} \ge 1\,$. Then by the root mean square inequality $\,\sqrt{\dfrac{1}{3}\left(\dfrac{9p^2}{s^2}+\dfrac{9q^2}{s^2}+\dfrac{9r^2}{s^2}\right)} \ge \dfrac{1}{3}\left(\dfrac{3p}{s}+\dfrac{3q}{s}+\dfrac{3r}{s}\right) \ge 1\,$ with equality iff $\,p=q=r=\dfrac{s}{3}\,$ which corresponds to a degenerate quadrilateral where two of the vertices divide the edge between the other two in equal parts. So in the end $\,\dfrac{p^2+q^2+s^2}{s^2} \ge \dfrac{1}{3}\,$, with strict inequality if degenerate quadrilaterals are excluded.

Answer 2

Euler’s theorem states that for $p+q+r\ge s$. Hence, to minimise the total value, we have $p=q=r=\frac{1}{3}s$.

$\frac{3\times\frac{1}{9}s^2}{s^2}=\frac{1}{3}$.