m-3z》0 加上 4z-m》0 得出z》0，z都大于0了，m肯定大于0，所以m+z大于0

(1) m - 3z > 0
(2) 4z - m > 0
1+2 z>0

条件（1）和（2）得出，4z>m>3z,即：4z>3z, 可否得出 Z>0 ? 故，得出 z+m 必定 大于0?

mark。不等式变形及大小比较。

条件1+2：首先将m - 3z > 0和4z - m > 0两个式子相加，得到z>0，由于由上面的两个式子得到m+z>4z，得到m+z>0.充分。

1+2， z>0, 再由1， m>z>0 , suf

(1) m - 3z > 0. Insufficient on its own.
(2) 4z - m > 0. Insufficient on its own.

(1)+(2) Remember we can add inequalities with the sign in the same direction --> \(m-3z+4z-m>0\) --> \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) --> \(m+z=positive+positive>0\). Sufficient.

(1)+(2)：3z0，z+m>0.

两个条件结合，3z