Let A=a bc d, where a,b,c,d∈R Then
det(A)≤a2+b2c2+d2
det(A)≤(a+b)(c+d)
det(A)≤ac+bd
det(A)≤(|a|−|b|)(|c|−|d|)
det(A)=ad−bc≤|ad−bc|≤|a∥d|+|b∥c|
Now,
a2+b2c2+d2−(|a∥d|+|b∥c|)2=a2c2+b2c2+a2d2+b2d2−a2d2+b2c2+2|a∥d∥b∥c|=(|ac|−|bd|)2≥0
⇒|a∥d|+|b||c|≤a2+b2c2+d2