2xtan−1x>log(1+x2)2x\tan^{-1}x > \log(1+x^2)2xtan−1x>log(1+x2) 2xtan−1x−log(1+x2)>02x\tan^{-1}x - \log(1+x^2) > 02xtan−1x−log(1+x2)>0 であればよい tan−1x−12xlog(1+x2)=f(x)\tan^{-1}x - \frac{1}{2x}\log(1+x^2) = f_{(x)}tan−1x−2x1log(1+x2)=f(x) とおくと f(x)′=1x2+1+12x2log(1+x2)−2x2x(1+x2)=12x2log(1+x2)f_{(x)}{}' = \frac{1}{x^2+1} + \frac{1}{2x^2}\log(1+x^2) - \frac{2x}{2x(1+x^2)} = \frac{1}{2x^2}\log(1+x^2)f(x)′=x2+11+2x21log(1+x2)−2x(1+x2)2x=2x21log(1+x2) x>0x > 0x>0 について f(x)′>0f_{(x)}{}' > 0f(x)′>0 となる x0・・・f(x)′f_{(x)}{}'f(x)′+f(x)f_{(x)}f(x)0 これから 2xtan−1x>log(1+x2)2x\tan^{-1}x > \log(1+x^2)2xtan−1x>log(1+x2) が成り立つ