シャピロの不等式
また、正の数 a, b, c, d に対して、相加平均と相乗平均の不等式から、 b a + c b + d c + a d ≥ 4 b a c b d c a d 4 = 4 {\displaystyle {\frac {b}{a}}+{\frac {c}{b}}+{\frac {d}{c}}+{\frac {a}{d}}\geq 4{\sqrt[{4}]{{\frac {b}{a}}{\frac {c}{b}}{\frac {d}{c}}{\frac {a}{d}}}}=4}
ここで S 4 := x 1 x 2 + x 3 + x 2 x 3 + x 4 + x 3 x 4 + x 1 + x 4 x 1 + x 2 {\displaystyle S_{4}:={\frac {x_{1}}{x_{2}+x_{3}}}+{\frac {x_{2}}{x_{3}+x_{4}}}+{\frac {x_{3}}{x_{4}+x_{1}}}+{\frac {x_{4}}{x_{1}+x_{2}}}} とおくと 2 S 4 = x 1 + x 2 x 2 + x 3 + x 2 + x 3 x 3 + x 4 + x 3 + x 4 x 4 + x 1 + x 4 + x 1 x 1 + x 2 + x 3 x 2 + x 3 + x 4 x 3 + x 4 + x 1 x 4 + x 1 + x 2 x 1 + x 2 − 4 + S 4 ≥ 4 + x 3 x 2 + x 3 + x 4 x 3 + x 4 + x 1 x 4 + x 1 + x 2 x 1 + x 2 − 4 + S 4 = x 3 x 2 + x 3 + x 4 x 3 + x 4 + x 1 x 4 + x 1 + x 2 x 1 + x 2 + x 1 x 2 + x 3 + x 2 x 3 + x 4 + x 3 x 4 + x 1 + x 4 x 1 + x 2 = ( x 1 + x 3 ) ( 1 x 2 + x 3 + 1 x 4 + x 1 ) + ( x 2 + x 4 ) ( 1 x 3 + x 4 + 1 x 1 + x 2 ) ≥ 4 ( x 1 + x 3 ) x 1 + x 2 + x 3 + x 4 + 4 ( x 2 + x 4 ) x 1 + x 2 + x 3 + x 4 = 4 {\displaystyle {\begin{aligned}2S_{4}&={\frac {x_{1}+x_{2}}{x_{2}+x_{3}}}+{\frac {x_{2}+x_{3}}{x_{3}+x_{4}}}+{\frac {x_{3}+x_{4}}{x_{4}+x_{1}}}+{\frac {x_{4}+x_{1}}{x_{1}+x_{2}}}\\&+{\frac {x_{3}}{x_{2}+x_{3}}}+{\frac {x_{4}}{x_{3}+x_{4}}}+{\frac {x_{1}}{x_{4}+x_{1}}}+{\frac {x_{2}}{x_{1}+x_{2}}}-4+S_{4}\\&\geq 4+{\frac {x_{3}}{x_{2}+x_{3}}}+{\frac {x_{4}}{x_{3}+x_{4}}}+{\frac {x_{1}}{x_{4}+x_{1}}}+{\frac {x_{2}}{x_{1}+x_{2}}}-4+S_{4}\\&={\frac {x_{3}}{x_{2}+x_{3}}}+{\frac {x_{4}}{x_{3}+x_{4}}}+{\frac {x_{1}}{x_{4}+x_{1}}}+{\frac {x_{2}}{x_{1}+x_{2}}}\\&+{\frac {x_{1}}{x_{2}+x_{3}}}+{\frac {x_{2}}{x_{3}+x_{4}}}+{\frac {x_{3}}{x_{4}+x_{1}}}+{\frac {x_{4}}{x_{1}+x_{2}}}\\&=(x_{1}+x_{3})\left({\frac {1}{x_{2}+x_{3}}}+{\frac {1}{x_{4}+x_{1}}}\right)+(x_{2}+x_{4})\left({\frac {1}{x_{3}+x_{4}}}+{\frac {1}{x_{1}+x_{2}}}\right)\\&\geq {\frac {4(x_{1}+x_{3})}{x_{1}+x_{2}+x_{3}+x_{4}}}+{\frac {4(x_{2}+x_{4})}{x_{1}+x_{2}+x_{3}+x_{4}}}\\&=4\end{aligned}}}