$$
\eqalign{
& {\cal (}{\cal 杨}{\cal 志}{\cal 明}{\cal )}{\cal 设}a,b,c,d{\cal 为}{\cal 正}{\cal 实}{\cal 数}{\cal 且}abcd \geqslant 1{\text{ }}{\cal 求}{\cal 证}{\cal :} \cr
& (a + b)(b + c)(c + d)(d + a) \geqslant {\text{ }}(a + 1)(b + 1)(c + 1)(d + 1){\text{ }} \cr
& {\cal 证}{\cal 明}{\cal :} \cr
& {\cal 左}{\cal 边}{\text{ = }}\sum {d^2 } (ab + bc + ca) + a^2 c^2 + b^2 d^2 + 2abcd \cr
& {\cal 右}{\cal 边}{\text{ = }}\sum {abc} + \sum {ab + } \sum {a + ac + bd + 1 + abcd} {\text{ }} \cr
& {\cal 易}{\cal 证} a^2 c^2 + b^2 d^2 \geqslant ac + bd{\cal ,}{\cal 利}{\cal 用}{\cal 米}{\cal 尔}{\cal 黑}{\cal 德}{\cal 可}{\cal 证}{\cal :} \cr
& \sum {d^2 } (ab + bc + ca) \geqslant {\text{3}}\sum {abc} \cr
& \sum {d^2 } (ab + bc + ca) \geqslant {\text{3}}\sum a \cr
& {\cal 只}{\cal 需}{\cal 证}\sum {d^2 } (ab + bc + ca) \geqslant {\text{3}}\sum {ab} \cr
& \sum {d^2 } (ab + bc + ca) - {\text{3}}\sum {ab} = \cr
& (ac + bd - 2)\sum {ab} + \frac{1}
{2}\sum {ab{\text{(}}bc + ad - 2{\text{)}}} \geqslant 0 \cr
& {\cal 万}{\cal 惠}{\cal 华}{\text{20181111}}\,\,\, \cr}
$$
|