Áp dụng BĐT phụ \(a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\)

\(A\ge\dfrac{1}{2}\left(x+y+\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+y+\dfrac{4}{x+y}\right)^2=\dfrac{1}{2}\left(1+\dfrac{4}{1}\right)^2=\dfrac{25}{2}\)

Dấu "=" \(x=y=\dfrac{1}{2}\)

a: \(\left(x-1\right)^2+\left(y-1\right)^2-2xy\)

\(=x^2-2xy+y^2-2\left(x+y\right)+2\)

\(=\left(x-y\right)^2-2\left(x-y\right)+2\)

\(=2^2-2\cdot2+2=2\)

b: \(\frac{x-1}{2}=y\)

=>x-1=2y

=>x-2y=1

\(x^3-8y^3-6xy\)

\(=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-6xy\)

\(=x^2+2xy+4y^2-6xy\)

\(=x^2-4xy+4y^2=\left(x-2y\right)^2=1\)

\(x+\frac{1}{x}=y+\frac{1}{y}\)

=>\(x-y+\frac{1}{x}-\frac{1}{y}=0\)

=>\(\left(x-y\right)+\frac{y-x}{xy}=0\)

=>\(\left(x-y\right)\left(1-\frac{1}{xy}\right)=0\)

=>\(1-\frac{1}{xy}=0\)

=>xy=1

\(\frac{x}{x^2+1}+\frac{y}{y^2+1}=\frac{x}{x^2+xy}+\frac{y}{y^2+xy}\)

\(=\frac{x}{x\left(x+y\right)}+\frac{y}{y\left(x+y\right)}=\frac{1}{x+y}+\frac{1}{x+y}=\frac{2}{x+y}\)

\(\frac{2\left(x+y\right)}{x^2+y^2+2}=\frac{2\left(x+y\right)}{x^2+y^2+2xy}=\frac{2\left(x+y\right)}{\left(x+y\right)^2}=\frac{2}{x+y}\)

Do đó: \(\frac{x}{x^2+1}+\frac{y}{y^2+1}=\frac{2\left(x+y\right)}{x^2+y^2+2}\)

\(P=\sum\dfrac{1}{x+y+1}\ge\dfrac{9}{2\left(x+y+z\right)+3}=\dfrac{9}{2.1+3}=\dfrac{9}{5}\)

Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)

Công thức tổng quát:

Áp dụng vào bài toán thì ta có Q=0.75

\(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=2\left(1\right)\)

\(\Leftrightarrow y+\sqrt{y^2+1}=2\sqrt{x^2+1}-2x\left(2\right)\)

Mặt khác: \(\left(1\right)\Leftrightarrow x+\sqrt{x^2+1}=2\sqrt{y^2+1}-2y\left(3\right)\)

Cộng vế theo vế \(\left(2\right);\left(3\right)\) ta được: \(x=-y\)

Khi đó: \(Q=x\sqrt{y^2+1}+y\sqrt{x^2+1}=-y\sqrt{y^2+1}+y\sqrt{y^2+1}=0\)