>>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?