f(x)=2x方+2/根号下4x方+1
=2x^2+2/√(4x^2+1)
设√(4x^2+1)=t,则x^2=(t^2-1)/4,且t≥1,
所以
f(x)= 2x^2+2/√(4x^2+1)
=(t^2-1)/2+2/t
=t^2/2+2/t-1/2
=t^2/2+1/t+1/t -1/2……利用基本不等式可得下式
≥3 (t^2/2*1/t*1/t)^(1/3)-1/2
=3 (1/2)^(1/3) -1/2
当t^2/2=1/t时取到最小值,此时t=2^(1/3).