Ta có x²+y² / x-y = x²-2xy+y²+2xy / x-y

                            = (x-y)²+2xy / x-y

Mà xy = 1 => 2xy = 2. Thay vào, ta có

(x-y)²+2xy / x-y = (x-y)²+2 / x-y = (x-y)² / x-y + 2 / x-y

                                                  = x-y + 2 / x-y

Áp dụng BĐT Cauchy, ta có

x-y + 2 / x-y ≥ 2.√(x-y).2 / x-y] = 2.√2 = (√2)³

Vậy Min A = (√2)³

\(\left\{{}\begin{matrix}x+y=2m-1\left(1\right)\\x^2+y^2=m^2+2m-3\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\left(x+y\right)^2-2xy=m^2+2m-3\)

\(\Leftrightarrow\left(2m-1\right)^2-m^2-2m+3=2xy\)

\(\Leftrightarrow2xy=3m^2-6m+4\)

\(P_{min}\Leftrightarrow3m^2-6m+4\left(min\right)\)

\(3\left(m^2-2m+\dfrac{4}{3}\right)=3\left(m^2-2m+1+\dfrac{1}{3}\right)=3\left[\left(m-1\right)^2+\dfrac{1}{3}\right]=3\left(m-1\right)^2+1\ge1\)

\("="\Leftrightarrow m=1\)

Đặt \(\left\{{}\begin{matrix}x+\sqrt{x^2+1}=a>0\\y+\sqrt{y^2+1}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2+1}=a-x\\\sqrt{y^2+1}=b-y\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2ax=a^2-1\\2by=b^2-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{a^2-1}{2a}\\y=\dfrac{b^2-1}{2b}\end{matrix}\right.\)

 \(\Rightarrow\left(\dfrac{a^2-1}{2a}+\sqrt{\left(\dfrac{b^2-1}{2b}\right)+1}\right)\left(\dfrac{b^2-1}{2b}+\sqrt{\left(\dfrac{a^2-1}{2a}\right)+1}\right)=1\)

\(\Rightarrow\left(\dfrac{a^2-1}{2a}+\dfrac{b^2+1}{2b}\right)\left(\dfrac{b^2-1}{2b}+\dfrac{a^2+1}{2a}\right)=1\)

\(\Rightarrow\left(\dfrac{a+b}{2}+\dfrac{a-b}{2ab}\right)\left(\dfrac{a+b}{2}-\dfrac{a-b}{2ab}\right)=\dfrac{4ab}{4ab}=\dfrac{\left(a+b\right)^2}{4ab}-\dfrac{\left(a-b\right)^2}{4ab}\)

\(\Rightarrow\dfrac{\left(a+b\right)^2}{4}-\dfrac{\left(a+b\right)^2}{4ab}-\dfrac{\left(a-b\right)^2}{4\left(ab\right)^2}+\dfrac{\left(a-b\right)^2}{4ab}=0\)

\(\Rightarrow\dfrac{\left(a+b\right)^2}{4}\left(1-\dfrac{1}{ab}\right)+\dfrac{\left(a-b\right)^2}{4ab}\left(1-\dfrac{1}{ab}\right)=0\)

\(\Rightarrow\left(1-\dfrac{1}{ab}\right)\left(\dfrac{\left(a+b\right)^2}{4}+\dfrac{\left(a-b\right)^2}{4ab}\right)=0\)

\(\Rightarrow1-\dfrac{1}{ab}=0\Rightarrow ab=1\)

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

\(\Rightarrow x+y=0\Rightarrow y=-x\)

\(P=2\left(x^2+\left(-x\right)^2\right)+0=4x^2\ge0\)

Dấu "=" xảy ra khi \(x=y=0\)

+Tìm điều kiện để hệ có nghiệm: 

\(\left(x-y\right)^2\ge0\Rightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)

\(\Rightarrow2\left(m^2+2m-3\right)\ge\left(2m-1\right)^2\)

\(\Leftrightarrow-2m^2+8m-7\ge0\)

\(\Leftrightarrow\frac{4-\sqrt{2}}{2}\le m\le\frac{4+\sqrt{2}}{2}\)

+Tìm m để xy nhỏ nhất:

\(xy=\frac{\left(x+y\right)^2-\left(x^2+y^2\right)}{2}=\frac{\left(2m-1\right)^2-\left(m^2+2m-3\right)}{2}=\frac{3}{2}\left(m^2-2m\right)+2\)

\(=\frac{3}{2}\left(m-1\right)^2+\frac{1}{2}\)

Để xy nhỏ nhất thì \(\left(m-1\right)^2\)phải nhỏ nhất;

\(m\ge\frac{4-\sqrt{2}}{2}\approx1,29\)

\(\Rightarrow m-1\ge\frac{4-\sqrt{2}}{2}-1=1-\frac{\sqrt{2}}{2}>0\)

\(\Rightarrow\left(m-1\right)^2\ge\left(1-\frac{\sqrt{2}}{2}\right)^2\)

Dấu bằng xảy ra khi \(m=\frac{4-\sqrt{2}}{2}\)

Đây là giá trị m cần tìm

\(1,\) Áp dụng BĐT: \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\text{ và }\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

Dấu \("="\Leftrightarrow x=y\)

\(A=\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2+17\ge\dfrac{1}{2}\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2+17\\ A\ge\dfrac{1}{2}\left(1+\dfrac{1}{a}+\dfrac{1}{b}\right)^2+17\ge\dfrac{1}{2}\left(1+\dfrac{4}{a+b}\right)^2+17=\dfrac{25}{2}+17=\dfrac{59}{2}\\ \text{Dấu }"="\Leftrightarrow\left\{{}\begin{matrix}a+\dfrac{1}{a}=b+\dfrac{1}{b}\\a+b=1\end{matrix}\right.\Leftrightarrow a=b=\dfrac{1}{2}\)

\(2,\text{Đặt }A=\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2\left(\dfrac{xy^2z}{xz}+\dfrac{xyz^2}{xy}+\dfrac{x^2yz}{yz}\right)\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2\left(x^2+y^2+z^2\right)\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+6\)

Áp dụng Cosi: \(\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}\ge2y^2\)

CMTT: \(\left\{{}\begin{matrix}\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}\ge2z^2\\\dfrac{x^2y^2}{z^2}+\dfrac{x^2z^2}{y^2}\ge2x^2\end{matrix}\right.\)

Cộng VTV \(\Leftrightarrow A^2\ge2\left(x^2+y^2+z^2\right)+6=12\\ \Leftrightarrow A\ge2\sqrt{3}\)

Dấu \("="\Leftrightarrow x=y=z=1\)