Giả thiết có: abc+bca+cda+dab = a+b+c+d+\(\sqrt{2012}\)
\(\Leftrightarrow\) (abc+bca+cda+dab-a-b-c-d)2 =2012
\(\Leftrightarrow\) \(\left[\left(abc-c\right)+\left(dab-d\right)+\left(bcd-b\right)+\left(cda-a\right)\right]^2\) = 2012
\(\Leftrightarrow\) \(\left[c\left(ab-1\right)+d\left(ab-1\right)+b\left(cd-1\right)+a\left(cd-1\right)\right]^2\) = 2012
\(\Leftrightarrow\) \(\left[\left(ab-1\right)\left(c+d\right)+\left(ab-1\right)\left(a+b\right)\right]^2\) = 2012
Áp dụng BĐT Bunhia cho 2 cặp số: (ab-1 ; a+b);(cd-1 ; c+d)
Ta có: \(\left[\left(ab-1\right)\left(c+d\right)+\left(ab-1\right)\left(a+b\right)\right]^2\) \(\le\) \(\left[\left(ab-1\right)^2+\left(a+b\right)^2\right]\left[\left(cd-1\right)^2+\left(c+d\right)^2\right]\)
\(\Leftrightarrow\) 2012 \(\le\) ( a2b2-2ab+1+a2+2ab+b2) (c2d2-2cd+1+c2+2cd+d2)
\(\Leftrightarrow\) 2012\(\le\) ( a2b2 +a2+b2+1)(c2d2+c2+d2+1)
\(\Leftrightarrow\) 2012 \(\le\) (a2+1)(b2+1)(c2+1)(d2+1) (đpcm)
