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

1. \(1=x^2+y^2\ge2xy\Rightarrow xy\le\frac{1}{2}\)

 \(A=-2+\frac{2}{1+xy}\ge-2+\frac{2}{1+\frac{1}{2}}=-\frac{2}{3}\)

max A = -2/3 khi x=y=\(\frac{\sqrt{2}}{2}\)

\(\frac{1}{xy}+\frac{1}{xz}=\frac{1}{x}\left(\frac{1}{y}+\frac{1}{z}\right)\ge\frac{1}{x}.\frac{4}{y+z}=\frac{4}{\left(4-t\right)t}=\frac{4}{4-\left(t-2\right)^2}\ge1\) với t = y+z => x =4 -t

\(A=x^2+y^2=\frac{\left(1^2+1^2\right)\left(x^2+y^2\right)}{2}\ge\frac{\left(1.x+1.y\right)^2}{2}=\frac{1}{2}\)A min = 1 khi x =y = 1/2

\(\sqrt{A}=\sqrt{x^2+y^2}\le\sqrt{x^2}+\sqrt{y^2}=x+y=1\)( \(\sqrt{a+b}\le\sqrt{a}+\sqrt{b}\))

=> A\(\le1\) => Max A = 1 khi x =0;y =1 hoặc x =1 ; y =0

Ta có \(x^2+y^2\ge2xy\)=>\(xy\le\frac{1}{2}\)

\(\frac{1}{A}=\frac{1}{-2xy}-\frac{1}{2}\le-1-\frac{1}{2}=-\frac{3}{2}\)

=> \(A\ge-\frac{2}{3}\)

\(MinA=-\frac{2}{3}\)khi \(x=y=\frac{\sqrt{2}}{2}\)

Trần Phúc Khang: bài này cần gì phải làm phức tạp vậy a

c/m: \(xy\le\frac{1}{2}\)( như bài Trần Phúc Khang)

Dấu "=" xảy ra <=> x=y=\(\frac{1}{\sqrt{2}}\)

\(A=\frac{-2xy}{1+xy}\ge\frac{-2.\frac{1}{2}}{1+\frac{1}{2}}=-\frac{1}{\frac{3}{2}}=-\frac{2}{3}\)

Dấu "=" xảy ra <=> x=y=\(\frac{1}{\sqrt{2}}\)

KL:.............................

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

\(A_{min}=-1\) khi \(2x+y=0\)

\(x+y=1\Rightarrow y=1-x\)

\(P=x^3+\left(1-x\right)^3+x\left(1-x\right)\)

\(P=2x^2-2x+1=\dfrac{1}{2}\left(2x-1\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\)

\(P_{min}=\dfrac{1}{2}\) khi \(x=y=\dfrac{1}{2}\)

1/a/
\(A=\frac{2}{xy}+\frac{3}{x^2+y^2}=\left(\frac{1}{xy}+\frac{1}{xy}+\frac{4}{x^2+y^2}\right)-\frac{1}{x^2+y^2}\)

\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}-\frac{1}{\frac{\left(x+y\right)^2}{2}}=16-2=14\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

b/

\(4B=\frac{4}{x^2+y^2}+\frac{8}{xy}+16xy=\left(\frac{4}{x^2+y^2}+\frac{1}{xy}+\frac{1}{xy}\right)+\left(\frac{1}{xy}+16xy\right)+\frac{5}{xy}\)

\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}+2\sqrt{\frac{1}{xy}.16xy}+\frac{5}{\frac{\left(x+y\right)^2}{4}}\)

\(=16+8+20=44\)

\(\Rightarrow B\ge11\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

2/

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

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

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)