Volume Inequality in Tetrahedron

Problem

Volume Inequality in Tetrahedron, problem

Solution 1

Expressing the volume in two ways, we get $abc=xbc+yca+zab.$ By the AM-GM inequality, $(xbc+yca+zab)^3\ge 27xyz(abc)^2.$ Thus, $(abc)^3\ge 27xyz(abc)^2,$ from which $abc\ge 27xyz.$

Equality is attained when $M$ is the centroid of $\Delta ABC.$ Indeed, let $O=(0,0,0),$ $A=(a,0,0),$ $B=(0,b,0)$ and $C=(0,0,c).$ Then $M=(ma,nb,pc)$ with $m+n+p=1.$ Equality holds iff $[MOAB]=[MOBC]=[MOCA]$ i.e. $mabc=nabc=pabc,$ from which $\displaystyle m=n=p=\frac{1}{3},$ making $M$ the centroid of $\Delta ABC.$

Solution 2

First we prove that

(*)

$\displaystyle \frac{x}{OA}+\frac{y}{OB}+\frac{z}{OC}=1.$

Indeed, writing the value of the tetrahedron $OABC$ in two ways, we obtain

$Vol(OABC)=Vol(MOAC)+Vol(MOAC)+Vol(MOAB),$

or, equivalently,

$\displaystyle \frac{OB\cdot OC\cdot x}{6}+\frac{OC\cdot OA\cdot x}{6}+\frac{OA\cdot OB\cdot x}{6}=\frac{OA\cdot OB\cdot OC}{6}.$

Dividing by the right-hand side proves (*). Applying to (*) the AM-GM inequality, we have

$\displaystyle 1=\frac{x}{OA}+\frac{y}{OB}+\frac{z}{OC}\ge 3\sqrt[3]{\frac{x}{OA}\cdot\frac{y}{OB}\cdot\frac{z}{OC}}$

which is equivalent to the require inequality.

Equality occurs, iff, $\displaystyle \frac{x}{OA}=\frac{y}{OB}=\frac{z}{OC}\;\left(=\frac{1}{3}\right),$ i.e., if the tetrahedron $OABC$ is similar to the tetrahedron with the apex at $M$ and the base formed by the three projections of $M$ on the faces of $OAB.$ In other words, if the two tetrahedra are homothetic with the coefficient $\displaystyle \frac{1}{3}.$ This only happens when $M$ is the centroid of $\Delta ABC.$

Solution 3

Let us choose the origin at $O$. WLOG, let $OA$, $OB$, and $OC$ be along the $X$, $Y$, and $Z$, respectively. The equation of the plane $ABC$ is $\displaystyle \frac{x}{OA}+\frac{y}{OB}+\frac{z}{OC}=1.$

The coordinate of point $M$ is $(x,y,z)$. As a result of the interior constraint, $x,y,z\geq 0$ and $\displaystyle \frac{x}{OA}+\frac{y}{OB}+\frac{z}{OC}\leq 1.$

Thus, from $AM-GM$,

$\displaystyle 3\left(\frac{xyz}{OA\cdot OB\cdot OC}\right)^{1/3}\leq 1 ~\Rightarrow~ 27xyz \leq OA\cdot OB\cdot OC. $

Acknowledgment

Dorin Marghidanu has kindly posted the problem at the CutTheKnotMath facebook page. He then commented with Solution 2, while Leo Giugiuc commented with Solution 1. Solution 3 is by Amit Itagi.

 

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