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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 ]