シャピロの不等式
よって、 S 3 := x 1 x 2 + x 3 + x 2 x 3 + x 1 + x 3 x 1 + x 2 {\displaystyle S_{3}:={\frac {x_{1}}{x_{2}+x_{3}}}+{\frac {x_{2}}{x_{3}+x_{1}}}+{\frac {x_{3}}{x_{1}+x_{2}}}} とおくと 2 S 3 = x 3 + x 1 x 2 + x 3 + x 1 + x 2 x 2 + x 3 + x 1 + x 2 x 3 + x 1 + x 2 + x 3 x 3 + x 1 + x 2 + x 3 x 1 + x 2 + x 3 + x 1 x 1 + x 2 − 3 ≥ 2 + 2 + 2 − 3 = 3 {\displaystyle {\begin{aligned}2S_{3}&={\frac {x_{3}+x_{1}}{x_{2}+x_{3}}}+{\frac {x_{1}+x_{2}}{x_{2}+x_{3}}}+{\frac {x_{1}+x_{2}}{x_{3}+x_{1}}}\\&+{\frac {x_{2}+x_{3}}{x_{3}+x_{1}}}+{\frac {x_{2}+x_{3}}{x_{1}+x_{2}}}+{\frac {x_{3}+x_{1}}{x_{1}+x_{2}}}-3\\&\geq 2+2+2-3=3\end{aligned}}}
ゆえに S 3 ≥ 3 2 {\displaystyle S_{3}\geq {\frac {3}{2}}} 。
n = 4 {\displaystyle n=4}
正の数 a, b に対して、相加平均と調和平均の不等式から、 1 a + 1 b ≥ 4 a + b {\displaystyle {\frac {1}{a}}+{\frac {1}{b}}\geq {\frac {4}{a+b}}}