Có 1/(1+a^2)+1/ (1+b^2)>= 2/(1+ab)
<=> 1/(1+a^2)+1/ (1+b^2) -2/(1+ab)>= 0
<=> [1/(1+a^2) -1/ (1+ab)] +[1/ (1+b^2)- 1/(1+ab)] >=0
<=> (ab- a^2)/[(1+a^2)(1+ab)]+ (ab- b^2)/[(1+b^2)(1+ab)]>=0
<=> [(ab- a^2)(1+b^2)+ (ab- b^2)(1+a^2)]/ [(1+a^2)(1+ab)(1+b^2)]>=0
<=> (ab- a^2)(1+b^2)+ (ab- b^2)(1+a^2)>=0 vì (1+a^2)(1+ab)(1+b^2)>0
<=> ab+ ab^3-a^2-a^2b^2+ab+a^3b-b^2-a^2b^2>=0
<=> ab^3-2a^2b^2+a^3b-a^2-b^2+2ab>=0
<=> ab(a^2- 2ab+b^2)- (a^2- 2ab+b^2)>=0
<=> ab(a-b)^2- (a-b)^2>=0
<=> (a-b)^2(ab-1)>=0 (luôn đúng => đpcm)
Dấu "=" xảy ra <=> a=b=1
