>>8　に習って

π = 6∫[0 to 1/√3] dx/(1+x^2)
　≧ 6∫[0 to 1/√3] (1 -x^2 +x^4 -x^6) dx
　= 6 [ x -(1/3)x^3 +(1/5)x^5 -(1/7)x^7 ](x=0,1/√3)
　= 6 [ x{1 -(1/3)x^2 +(1/5)x^4 -(1/7)x^6} ](x=0,1/√3)
　= (2√3)(1 -1/9 +1/45 -1/189)
　= (2√3)(856/945)
　= 3.137852892　

　√3 = 1.732050808 は使ってもいいな？