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Arithmetic Mean – Geometric Mean Problem (Tomato subjective 82)

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 ]

July 31, 2015
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