a,

Có : 1/x + 1/y >= 4/x+y = 4/1 = 4

Dấu "=" xảy ra <=> x=y=1/2

Vậy ..............

b, Áp dụng bđt sovac ta có : 

a^2/x + b^2/y >= (a+b)^2/x+y = (a+b)^2 >= 0

Dấu "=" xảy ra <=> x=y=1/2 và a=-b

Vậy ..............

Tk mk nha

câu c áp dụng \(a^2+b^2\ge\frac{1}{2}\left(a+b\right)^2\) và \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)bạn tự giải nhá.

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

Áp dụng BĐT Bunhiacoxki có: 

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

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

Theo BĐT Cauchy thì: \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

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

=> \(A_{min}=\frac{25}{2}\)

Dấu "=" xảy ra khi x=y=1/2

\(a)\) Có \(2012=x+y\ge2\sqrt{xy}\)\(\Leftrightarrow\)\(xy\le1006^2\)

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

\(\le2+\frac{4.1006^2}{2012^2}=2\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=1006\)

\(b)\) \(C=\left(1+\frac{2012}{x}\right)^2+\left(1+\frac{2012}{y}\right)^2\ge\left[2+2012\left(\frac{1}{x}+\frac{1}{y}\right)\right]^2\ge\left(2+\frac{2012.4}{x+y}\right)^2\)

\(=\left(2+\frac{2012.4}{2012}\right)^2=36\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=1006\)

... 

cảm ơn bạn nhiều

TXD : \(\hept{\begin{cases}y\left(x+y\right)\ne0\\\left(x+y\right)x\ne0\\\left(x-y\right)\left(x+y\right)\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne y\\x\ne-y\\xy\ne0\end{cases}}}\)

Câu b :

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

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

Để \(A>1\)mà \(y< 0\)nên \(x\)và \(y\)phải cùng dấu \(\Rightarrow x< 0\)

Câu 2 thế y = 1 - x rồi quy đồng như bình thường là ra bn nhé

ta có ; A=((x+2012)/x)^2 + ((y+2012)/y)^2

  hay A  =((x+x+y)/x)^2+((y+x+y)/x)^2

            =((2x+y)/x)^2 + ((2x+y)/x)^2

            =(2+y/x)^2 + (2+x/y)^2

đặt x/y=k ta có ;

A=(2+k)^2 + (2+1/k)^2

  =4+4k+k^2+4+4/k+1/k^2 

   \(\ge\)\(2\sqrt{4k.\frac{1}{4k}}\)+\(2\sqrt{k^2.\frac{1}{k^2}}\)\(+8\)(\(BAT\)\(DANG\)\(THUC\)\(COSI\))

   \(=\)\(2\sqrt{1}+2\sqrt{16}+8=2+8+8=18\)

\(_{ }\)vậy max A = 18