\(x^3+y^3=\left(x+y\right).\left[\left(x+y\right)^2-3xy\right]\)\(=1-3xy\)
\(\Rightarrow A=\dfrac{1}{1-3xy}+\dfrac{1}{xy}\)
\(\Rightarrow A=\dfrac{1}{1-3xy}+\dfrac{3}{3xy}\ge\dfrac{\left(1+\sqrt{3}\right)^2}{1-3xy+3xy}\)
\(\Rightarrow A=4+2\sqrt{3}\)
Đẳng thức xảy ra \(\Leftrightarrow xy=\dfrac{3-\sqrt{3}}{6}\)
Thay \(1=\left(x+y\right)^3\) vào biểu thức \(A\) ta có:
\(A=\dfrac{\left(x+y\right)^3}{x^3+y^3}+\dfrac{\left(x+y\right)^3}{xy}\)
\(A=\dfrac{x^3+y^3+3xy.\left(x+y\right)}{x^3+y^3}+\dfrac{x^3+y^3+3xy.\left(x+y\right)}{xy}\)
\(A=1+\dfrac{3xy}{x^3+y^3}+3+\dfrac{x^3+y^3}{xy}\)
\(A=4+\left(\dfrac{3xy}{x^3+y^3}+\dfrac{x^3+y^3}{xy}\right)\ge4+2\sqrt{\dfrac{3xy.\left(x^3+y^3\right)}{xy.\left(x^3+y^3\right)}}\)
\(A=4+2\sqrt{3}\) (Áp dụng BĐT Cauchy cho 2 số dương)
\(\Rightarrow A_{min}=\left(\sqrt{3}+1\right)^2\) khi \(\dfrac{3xy}{x^3+y^3}=\dfrac{x^3+y^3}{xy}\)
\(\Leftrightarrow x^3+y^3=xy\sqrt{3}\)
\(\Leftrightarrow\left(x+y\right).\left(x^2-xy+y^2\right)=xy\sqrt{3}\)
\(\Leftrightarrow x^2+y^2-xy\left(\sqrt{3}+1\right)=0\) và \(x+y=1\).
Đến đây thay \(x=1-y\) vào pt trên ta có:
\(y^2.\left(3+\sqrt{3}\right)-y\left(3+\sqrt{3}\right)+1=0\) có:
\(\Delta=\left(3+\sqrt{3}\right)^2-4.\left(3+\sqrt{3}\right)=2\sqrt{3}\)
\(\Rightarrow\sqrt{\Delta}=\sqrt{2\sqrt{3}}\)
\(\Rightarrow y=\dfrac{3+\sqrt{3}+\sqrt{2\sqrt{3}}}{2.\left(3+\sqrt{3}\right)}\)
\(\Rightarrow x=1-y=\dfrac{3+\sqrt{3}-\sqrt{2\sqrt{3}}}{2.\left(3+\sqrt{3}\right)}\)
