Try this Arithmetic Mean - Geometric Mean Subjective Problem number 82 from TOMATO.
Problem: Arithmetic Mean - Geometric Mean
Let $ {a, b, c, d}$ be positive real numbers such that $ {abcd = 1}$. Show that,
$ {\displaystyle{(1 + a)(1 + b)(1 + c)(1 + d) {\ge} {16}}}$
Solution: $ {{\sum{a}} = a + b + c + d}$
= $ {\displaystyle{4 \left({\frac{a + b + c + d}{4}}\right)}}$
$ {\displaystyle{\ge}}$ $latex {\displaystyle{4 {(abcd)^{\frac{1}{4}}}}}$ [ AM-GM ]
= 4
$ {\displaystyle{\sum{ab}}}$ = $latex {\displaystyle{6 \left({\frac{\sum {ab}}{6}}\right)}}$
$ {\displaystyle{\ge}}$ $latex {\displaystyle{6 {(abcd)^{\frac{2}{6}}}}}$ [ AM-GM ]
= 6
$ {\displaystyle{\sum{abc}}}$ = $latex {\displaystyle{4 \left({\frac{\sum {abc}}{4}}\right)}}$
$ {\displaystyle{\ge}}$ $latex {\displaystyle{4 {(abcd)^{\frac{3}{4}}}}}$ [ AM-GM ]
= 4
Now, L.H.S = $ {\displaystyle{(1 + a)(1 + b)(1 + c)(1 + d)}}$
= $ {\displaystyle{1 + abcd + {\sum{a}} + {\sum{ab}} + {\sum{abc}}}}$
$ {\displaystyle{\ge}}$ $latex {\displaystyle{1 + 1 + 4 + 6 + 4}}$
= 16
= RHS. [ proved ]
