Is xy+zt+yz+tx > 0?

\(\underline {xy} + \underline{\underline {zt}} + \underline {yz} + \underline{\underline {tx}} \,\,\mathop > \limits^? \,\,0\,\,\,\,\, \Leftrightarrow \,\,\,\,\,y\left( {x + z} \right) + t\left( {z + x} \right) = \left( {x + z} \right)\left( {y + t} \right)\,\,\,\mathop > \limits^? \,\,0\,\,\,\,\,\left( * \right)\,\)


\(\left( 1 \right)\,\,\,\left| x \right| = \left| y \right| = \left| z \right| = t\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {x,y,z,t} \right) = \left( {0,0,0,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {x,y,z,t} \right) = \left( {1,1,1,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)

\(\left( 2 \right)\,\,\,x + y + z + t = 0\,\,\,\, \Rightarrow \,\,\,\,x + z = - \left( {y + t} \right)\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,?\,\,\,:\,\,\,\left( {x + z} \right)\left( {y + t} \right) = - {\left( {y + t} \right)^2} \leqslant 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle\)


The correct answer is (B).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.