i) WLOG $ a \ge b \ge c \ge d$. Then $ \frac {1}{2}(c + d) \le td + cz + (1 - t - d)b \le td + cz + by + ax$ ii) It suffices to show that $ \frac {c + d}{2} \ge 54abcd$. It then suffices to show that $ c + d \ge 108(\frac {a + b}{2})^2(\frac {c + d}{2})^2$ $ \leftrightarrow 4(c + d) \ge 27(1 - (c + d))^2(c + d)^2$ $ \leftrightarrow 4 \ge 27(1 - (c + d))^2(c + d)^2$ Which is true since $ 27((1 - (c + d))(c + d))^2 \le 27(\frac {c + d + (1 - (c + d))}{2})^4 = \frac {27}{16} < 4$ The stronger inequality $ ax + by + cz + dt \ge 54(\sqrt {ab})cd$, was probably intended, since then equality can actually be obtained (use the same method as above). Cheers, Rofler