On the base of the $ABC$ of the triangular pyramid $SABC$ mark the point $M$ and through it were drawn lines parallel to the edges $SA, SB$ and $SC$, which intersect the side faces at the points $A1_, B_1$ and $C_1$, respectively. Prove that $\sqrt{MA_1}+ \sqrt{MB_1}+ \sqrt{MC_1}\le \sqrt{SA+SB+SC}$
Let $(p, q, r)$ be the barycentric coordinates of $M$ wrt $ABC$. Define $A_2=AM\cap BC, B_2 =BM\cap CA, C_2 = CM\cap AB$. Then
$$MA_1=SA\cdot \frac{MA_2}{AA_2}=SA\cdot \frac{p}{p+q+r}$$and the same is true for $MB_1$ and $MC_1$. Thus by C-S
$$\left(\sqrt{MA_1}+ \sqrt{MB_1}+ \sqrt{MC_1}\right)^2=\Bigg(\sqrt{SA\cdot \frac{p}{p+q+r}}+ \sqrt{SB\cdot \frac{q}{p+q+r}}+ \sqrt{SC\cdot \frac{r}{p+q+r}}\Bigg)^2\le$$$$\le\left(\frac{p}{p+q+r}+\frac{q}{p+q+r}+\frac{r}{p+q+r}\right)(SA+SB+SC)=SA+SB+SC.$$