8ab(1-ab)<=(1+a^2)(1+b^2) <=> 8ab<=1+a^2+b^2+9a^2*b^2 ...(*)
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Avem a^2+b^2=|a|^2+|b|^2>=2*|a|*|b|>=2ab si 1+9a^2*b^2>=2*rad(1*9*a^2*b^2)=2*3*|a|*|b|=6|a|*|b|>=6ab.
Deci (a^2+b^2)+(1+9a^2*b^2)>=2ab+6ab=8ab, ceea ce-i echivalent cu (*).
Observatie: Egalitatea are loc daca si numai daca a=b=1/rad(3) sau a=b=-1/rad(3).
