Cho $a,b,c,d$ là các số thực thỏa mãn điều kiện: $abc+bcd+cda+dab=a+b+c+d+\sqrt{2012}$. CMR: $\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\left(d^2+1\right)\ge2012$

Bùi Ngọc Diễm

Ta có: $2012=\left(abc+bcd+cda+dab-a-b-c-d\right)^2$

$=\left\lbrack\left(ab-1\right)\left(c+d\right)+\left(cd-1\right)\left(a+b\right)\right\rbrack^2$

$\le\left\lbrack\left(ab-1\right)^2+\left(a+b\right)^2\right\rbrack\left\lbrack\left(cd-1\right)^2+\left(c+d\right)^2\right\rbrack$

$=\left(a^2b^2+a^2+b^2+1\right)\left(c^2d^2+c^2+d^2+1\right)$

$=\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\left(d^2+1\right)$

Vậy $\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\left(d^2+1\right)\ge2012$