ta co: a/(1+b²)=(a+ba²-ab²)/(1+b²)=(a(1+b²)-a...
Tuong tu: b/(1+c²)>=b-bc/2; c/(1+a²)>=c-ac/2.
=> a/(1+b²)+b/(1+c²)+c/(1+a²)>=a+b+c-1/2(ab...
Ma: 3(ab+bc+ca)<=(a+b+c)²=9=> ab+bc+ca <=3
=>-1/2(ab+bc+ca)>=-3/2
=> a+b+c-1/2(ab+bc+ca) >=3-3/2=3/2
=> a/(1+b²)+b/(1+c²)+c/(1+a²)>= 3/2(dpcm)
Dau "=" say ra <=> a=b=c=1