设a、b、c为正数,且a^2+b^2+c^2=3,证明:1/(1+2ab)+1/(1+2bc)+1/(1+2ca)>=1

设a、b、c为正数,且a^2+b^2+c^2=3,证明:1/(1+2ab)+1/(1+2bc)+1/(1+2ca)>=1.
a、b、c为正数,故
1/(1+2ab)+1/(1+2bc)+1/(1+2ca)
≥1/(1+a^2+b^2)+1/(1+b^2+c^2)+1/(1+c^2+a^2)
≥3*{3/[(a^2+b^2)+(b^2+c^2)+(c^2+a^2)+3]}
=9/[3+2(a^2+b^2+c^2)]
=1
即1/(1+2ab)+1/(1+2bc)+1/(1+2ca)≥1.
如果你会，帮我细细讲讲好么，