A = ab + bc + cd < ab + ad + bc + cd = ( a + c ) ( b + d )
Áp dụng bất đẳng thức xy < (\(\frac{x+y}{2}\) )2 ta có
A = ( a+ c ) ( b+ d ) < ( \(\frac{a+c+b+d}{2}\) )2 = \(\frac{1}{4}\)
A = \(\frac{1}{4}\) \(\Leftrightarrow\) \(\begin{cases}a+c=\frac{1}{2}\\b+d=\frac{1}{2}\\ad=0\\a,b,c,d\ge0\end{cases}\)
Vậy max A = \(\frac{1}{4}\) khi a= b = \(\frac{1}{2}\) , c = d = 0