Identitate conditionata (own !).

[Virgil Nicula]

Sa se arate ca pentru orice \( \{b,c,d,x,y,z\}\subset\mathrm R \) exista implicatia

\( \left\|\ \begin{array}{c} (z + d)(b + c) = 2(bc + zd)\\\\\\\\
(b + d)(x + c) = 2(bd + xc)\\\\\\\\
(c + d)(y + b) = 2(cd + yb)\end{array}\ \right\|\ \Longrightarrow\ (z + d)(x + y) = 2(zd + xy) \) .

[Mateescu Constantin]

Din cele 3 relatii exprimam pe \( x,\ y \) si \( z \) in functie de \( b,\ c \) si \( d \):

\( x=\frac{2bd-c(b+d)}{b+d-2c},\ y=\frac{2cd-b(c+d)}{c+d-2b},\ z=\frac{2bc-d(b+c)}{b+c-2d} \) .