I think question (1) is not true.It must be like this:$ S_{\triangle BDF} = (1 - x)zy S_{\triangle ABC}$ I will solve the question (2): use the equation (1) thus,We need to prove : $ \sqrt[3]{(1-x)zy}+\sqrt[3]{x(1-y)(1-z)} \leq 1$ use Cauchy inequality: $ LHS \leq \frac{1-x+z+y}{3}+\frac{x+1-y+1-z}{3}=1=RHS$ done

Agree with tuandokim on (1). For the inequality, you know that 0 $\leq$ x,y,z $\leq$ 1. So we can write (1-x,x) = ($\frac{1}{2},\alpha$), (y,1-y) = ($\frac{1}{2},\beta$) and (z,1-z) = ($\frac{1}{2},\alpha$). So the inequality is now: $\sqrt[3]{(\frac{1}{2}+\alpha)(\frac{1}{2}+\beta)(\frac{1}{2}+\gamma)} + \sqrt[3]{(\frac{1}{2}-\alpha)(\frac{1}{2}-\beta)(\frac{1}{2}-\gamma)} \leq 1$ However, we can just apply AM-GM for each individual term on the lhs and get the desired expression. (You didn't really need to use that substitution, but I did it to make everything look pretty.

IMOSL 1988 Q12