How to prove that $a^2+b^2+c^2 \geqslant162d^2$ in pyramid

Question

Given a quadrilateral pyramid $S.ABCD$ with the following properties :the base $ABCD$ is a rectangle and $SA$ is perpendicular to the plane $(ABCD)$. Suppose that $G$ is the centroid of the triangle $SBC$ and let $d$ be the distance from $G$ to the plane $(SBD)$.Let $SB=a;BD=b$ and $SD=c$. How to prove that $a^2+b^2+c^2 \geqslant 162d^2$ ?

Answer 1

Let $AD=x$, $AB=y$ and $AS=z$.

Since the distance from $G$ to $(SBD)$ equal to $\frac{1}{3}$ of the distance from $C$ to $(SBD)$, which equal to the distance from $A$ to $(SBD)$, we see that $d$ it's $\frac{1}{3}$ of the altitude $h$ of the pyramid $ASBD$ (from the vertex $A$, of course).

$a=\sqrt{y^2+z^2}$, $b=\sqrt{x^2+y^2}$ and $c=\sqrt{x^2+z^2}$.

Thus, $$S_{\Delta SBD}=\sqrt{p(p-a)(p-b)(p-c)}=\frac{1}{4}\sqrt{2\sum_{cyc}(2a^2b^2-a^4)}=$$ $$=\frac{1}{4}\sqrt{\sum_{cyc}(2(x^2+y^2)(x^2+z^2)-(x^2+y^2)^2)}=\frac{1}{2}\sqrt{x^2y^2+x^2z^2+y^2z^2}.$$ Thus, $$V_{ASBD}=\frac{1}{3}\cdot\frac{1}{2}\sqrt{x^2y^2+x^2z^2+y^2z^2}\cdot h=\frac{h\sqrt{x^2y^2+x^2z^2+y^2z^2}}{6}.$$ In another hand, $$V_{ASBD}=\frac{1}{3}S_{\Delta ABD}\cdot AS=\frac{1}{6}xyz,$$ which says that $$h=\frac{xyz}{\sqrt{x^2y^2+x^2z^2+y^2z^2}},$$ $$d=\frac{xyz}{3\sqrt{x^2y^2+x^2z^2+y^2z^2}}$$ and we need to prove that: $$x^2+y^2+x^2+z^2+y^2+z^2\geq\frac{162x^2y^2z^2}{9(x^2y^2+x^2z^2+y^2z^2)}$$ or $$(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)\geq9x^2y^2z^2,$$ which is true by AM-GM: $$(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)\geq3\sqrt[3]{x^2y^2z^2}\cdot3\sqrt[3]{x^4y^4z^4}=9x^2y^2z^2.$$ Done!