Minimum value of sum of volumes

Question

If $\Delta_{1},\Delta_{2},\Delta_{3},\Delta_{4}$ be the areas of triangular faces of Tetrahedron and $h_{1},h_{2},h_{3},h_{4}$ be the corresponding altitudes of the tetrahedron, Then minimum values of $\sum_{\mathop{1\leq i<j\leq 4}}(\Delta_{i}h_{j}),$ Given volume of Tetrahedron is $5$ cubic unit.

what i try

Given $\displaystyle \frac{1}{6}\Delta_{i}h_{i}=V=5\Rightarrow \Delta_{i}h_{i}=30\cdots \cdots (1)$ for $i=1,2,3,4$

Now we have to minimize

$$\Delta_{1}(h_{2}+h_{3}+h_{4})+\Delta_{2}(h_{3}+h_{4})+\Delta_{3}h_{4}$$

How do i minimize it Help me please

Answer 1

Let $\Delta_4=\max\{\Delta_1,\Delta_2,\Delta_3,\Delta_4\}$.

Thus, $\Delta_4<\Delta_1+\Delta_2+\Delta_3$ and $$\sum_{1\leq i<j\leq4}\Delta_ih_j=15\left(\frac{\Delta_1+\Delta_2+\Delta_3}{\Delta_4}+\frac{\Delta_1}{\Delta_2}+\frac{\Delta_1}{\Delta_3}+\frac{\Delta_2}{\Delta_3}\right)>$$ $$>15\left(1+\frac{\Delta_1}{\Delta_2}+\frac{\Delta_1}{\Delta_3}+\frac{\Delta_2}{\Delta_3}\right)>15.$$ Since for $\Delta_1+\Delta_2+\Delta_3\rightarrow\Delta_4,$ $\Delta_1\rightarrow0^+$ and $\Delta_3\rightarrow\Delta_4$ we see that $\sum\limits_{1\leq i<j\leq4}\Delta_ih_j\rightarrow15,$ so $15$ is an infimum and the minimum does not exist.

Answer 2

For any tetrahedron $\mathcal{T}$, let $V$ be its volume, $\Delta_i$ and $h_i$ be its face areas and corresponding altitudes. Since $$\color{red}{3} V = \Delta_1 h_1 = \Delta_2 h_2 = \Delta_3 h_3 = \Delta_4 h_4$$ The expression we want to minimize equals to

$$f(\mathcal{T}) \stackrel{def} = \sum_{1 \le i < j \le 4} \Delta_i h_j = 3V \sum_{1 \le i < j \le 4} \frac{\Delta_i}{\Delta_j} = 3V\left(\frac{\Delta_1 + \Delta_2 + \Delta_3}{\Delta_4} + \frac{\Delta_1 + \Delta_2}{\Delta_3} + \frac{\Delta_1}{\Delta_2}\right) $$ The task at hand is to minimize $f(\mathcal{T})$ subject to the constraint the four numbers $\Delta_1, \Delta_2, \Delta_3, \Delta_3$ can be realized as the face area of a tetrahedron $\mathcal{T}$ with $V = 5$.

In general, given four positive numbers $A_1, A_2, A_3, A_4$, it can be realized as the face area of a non-degenerate tetrahedron when and only when they satisfy following 4 inequalities (for a proof, see this): $$\begin{cases} A_1 + A_2 + A_3 &> A_4\\ A_2 + A_3 + A_4 &> A_1\\ A_3 + A_4 + A_1 &> A_2\\ A_4 + A_1 + A_2 &> A_3 \end{cases}\tag{*1}$$ This implies

$$f(\mathcal{T}) > 3V \left(\frac{\Delta_1 + \Delta_2 + \Delta_3}{\Delta_4}\right) > 3V = 15\tag{*2}$$

This leads to $$\inf\limits_{V(\mathcal{T}) = 5}f(\mathcal{T}) \ge 15$$

For any $\epsilon \in (0,1)$, consider the four numbers $(A_1,A_2,A_3,A_4) = (\epsilon^2,\epsilon,1,1)$. It is easy to see they satisfies inequalities $(*1)$ and hence can be realized by as the face area of some non-degenerate tetrahedron. By suitable scaling, we can find a tetrahedron $\mathcal{T}_{\epsilon}$ with volume $5$ and

$$\Delta_1 : \Delta_2 : \Delta_3 : \Delta_4 = A_1 : A_2 : A_3 : A_4 = \epsilon^2 : \epsilon : 1 : 1$$ For such a tetrahedron, we have

$$f(\mathcal{T}_\epsilon) = 15\left(\frac{\epsilon^2 + \epsilon +1}{1} + \frac{\epsilon^2 + \epsilon}{1} + \frac{\epsilon^2}{\epsilon}\right) = 15(1+3\epsilon+\epsilon^2)$$ Form this, we can deduce

$$\inf_{V(\mathcal{T}) = 5} f(\mathcal{T}) \le \lim_{\epsilon\to 0} f(\mathcal{T}_\epsilon) = 15$$

As a result, the infimum of $\sum\limits_{1 \le i < \le j} \Delta_i h_i$ is $15$. Since the inequality in $(*2)$ is always strict, the minimum of the expression at hand doesn't exist.