Problem: 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}} {\displaystyle{4 {(abcd)^{\frac{1}{4}}}}} [ AM-GM ]
= 4

{\displaystyle{\sum{ab}}} = {\displaystyle{6 \left({\frac{\sum {ab}}{6}}\right)}}
{\displaystyle{\ge}} {\displaystyle{6 {(abcd)^{\frac{2}{6}}}}} [ AM-GM ]
= 6

{\displaystyle{\sum{abc}}} = {\displaystyle{4 \left({\frac{\sum {abc}}{4}}\right)}}
{\displaystyle{\ge}} {\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}} {\displaystyle{1 + 1 + 4 + 6 + 4}}
= 16
= RHS. [ proved ]