Nếu \(0< x+y< 1\)
Do \(0< x+y< 1\Rightarrow\left\{{}\begin{matrix}0< x< 1\\0< y< 1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2< x\\y^2< y\end{matrix}\right.\)
\(\Rightarrow1>x+y>x^2+y^2\)
\(\Rightarrow log_{x+y}\left(x^2+y^2\right)>1\) (trái với giả thiết)
\(\Rightarrow x+y>1\)
Khi đó ta có: \(log_{x+y}\left(x^2+y^2\right)\le1\Leftrightarrow x+y\ge x^2+y^2\ge\dfrac{1}{2}\left(x+y\right)^2\)
\(\Rightarrow x+y\le2\)
Đặt \(x+y=a\Rightarrow1< a\le2\)
\(A=6a^2-14a+44=6a^2-14a+4+40=2\left(a-2\right)\left(3a-1\right)+40\)
Do \(1< a\le2\Rightarrow\left\{{}\begin{matrix}a-2\le0\\3a-1>0\end{matrix}\right.\) \(\Rightarrow\left(a-2\right)\left(3a-1\right)\le0\)
\(\Rightarrow A\le40\)
\(A_{max}=40\) khi \(a=2\) hay \(x=y=1\)