If we denote by $a,b,c,d~\text{and}~x,y,z,t~$ the respective volumes and areas of the faces, then $$\sqrt{\left(a+b+c+d\right)\left(\frac{1}x+\frac{1}y+\frac{1}z+\frac{1}t\right)}\geq\sqrt{\frac{a}x}+\sqrt{\frac{b}y}+\sqrt{\frac{c}z}+\sqrt{\frac{d}t},$$true by CBS.

parmenides51 wrote: Let $x_1, x_2, x_3, x_4$ be the distances from an arbitrary point inside the tetrahedron to the planes of its faces, and let $h_1, h_2, h_3, h_4$ be the corresponding heights of the tetrahedron. Prove that $$\sqrt{h_1+h_2+h_3+h_4} \ge \sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}+\sqrt{x_4}$$ (Dmitry Nomirovsky) $$\sqrt{\sum_{i=1}^4h_i}=\sqrt{\sum_{i=1}^4\frac{\sum\limits_{k=1}^4S_kx_k}{S_i}}=\sqrt{\sum_{k=1}^4\frac{1}{S_k}\sum\limits_{k=1}^4S_kx_k}\geq\sum_{k=1}^4\sqrt{x_i}.$$

For the collection: Let $A_1\ldots A_{n+1}~$ be a $n-\text{simplex}~(n\geq2)~$ and let $M~$ be an interior point of the simplex. We denote by $h_i~\text{the heights of the simplex and by}~x_i~\text{the distances from M to the faces of the simplex, where}~i=1,\ldots,n+1.$ $$\text{Prove}~\sqrt{\sum_{i=i}^{n+1}{h_i}}\geq\sum_{i=i}^{n+1}{\sqrt{x_i}}.$$