a21+a22)(b21+b22)≥(a1b1+a2b2)2(a12+a22)(b12+b22)≥(a1b1+a2b2)2
→a21b21+a21b22+a22b21+a22b22≥a21b21+2a1b1a2b2+a22b22→a12b12+a12b22+a22b12+a22b22≥a12b12+2a1b1a2b2+a22b22
→a21b22+a22b21−2a1b1a2b2≥0→a12b22+a22b12−2a1b1a2b2≥0
→(a1b2−a2b1)2≥0→(a1b2−a2b1)2≥0 (đúng)
Vậy ta có đpc