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

Áp dụng BĐT AM-GM: \(A\ge2\sqrt{\frac{x}{4x}}+2\sqrt{\frac{y}{4y}}+\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)=2+\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)

Áp dụng BĐT Schwarz dạng Engel: \(A\ge2+\frac{1}{4}.\frac{4}{x+y}\ge3\) (Do \(x+y\le1\))

Vậy Min A = 3. Dấu "=" xảy ra <=> x=y=1/2

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

Đặt \(\frac{x}{y}=a\left(a>0\right)\)

vì x,y>0 áp dụng bđt cô si

\(x+\frac{1}{y}\ge2\sqrt{\frac{x}{y}}\) 

\(1\ge x+\frac{1}{y}\ge2\sqrt{\frac{x}{y}}\)

\(\frac{1}{4}\ge\frac{x}{y}\)

\(0< a\le\frac{1}{4}\)

Có A=\(a+\frac{1}{a}\left(với0< a\le\frac{1}{4}\right)\)

A=​\(16a+\frac{1}{a}-15a\)

a>0 cô si

A\(\ge2\sqrt{16a\cdot\frac{1}{a}}-15\cdot\frac{1}{4}=\frac{17}{4}\)

D=XR x=y=1/2

Các bất đẳng thức đúng : \(ab\le\frac{\left(a+b\right)^2}{4};\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

Áp dụng ta được :

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

\(=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{3}{2xy}\)

Ta có :

\(\frac{1}{x^2+y^2}+\frac{1}{2xy}\ge\frac{4}{x^2+y^2+2xy}=\frac{4}{\left(x+y\right)^2}\ge4\)

\(\frac{3}{2xy}\ge\frac{3}{2.\frac{\left(x+y\right)^2}{4}}=\frac{3}{2.\frac{1}{4}}=6\)

\(\Rightarrow A=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{3}{2xy}\ge4+6=10\)

Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)

Vậy \(A_{min}=10\) tại \(x=y=\frac{1}{2}\)

thangwd hdashdfjdfishjdf

Vì a,b>0

A\(\ge2\sqrt{\frac{1}{x}\cdot\frac{1}{y}}\cdot\sqrt{1+x^2y^2}\)

A\(\ge2\sqrt{\frac{1+x^2y^2}{xy}}\)

A\(\ge2\sqrt{\frac{1}{xy}+xy}\)

Đặt xy=a, a>0

Ta cs xy\(\le\frac{\left(x+y\right)^2}{4}\le\frac{1^2}{4}=\frac{1}{4}\)

ĐK 0<a<\(\frac{1}{4}\)

\(\Leftrightarrow A\ge2\sqrt{\frac{1}{a}+a}\)

A\(\ge2\sqrt{16a+\frac{1}{a}-15a}\)

a>0, áp dụng bđt cô si

\(A\ge2\sqrt{2\sqrt{16a\cdot\frac{1}{a}}-\frac{15}{4}}\)

A\(\ge\sqrt{17}\)

Dấu = x ra a=b=0.5 

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

Ta co:\(x+\frac{1}{x}=\left(\frac{1}{x}+4x\right)-3x\ge2\sqrt{\frac{1}{x}\cdot4x}-3x=4-3x\left(AM-GM\right)\)

Tuong tu:\(y+\frac{1}{y}=4-3y\)

Ta co:\(A\ge\left(4-3x\right)^2+\left(4-3y\right)^2\)

\(=16-24x+9x^2+16-24y+9y^2\)

\(=32-24\left(x+y\right)+9\left(x^2+y^2\right)\)

Ap dung bat dang thuc phu:\(\frac{\left(x+y\right)^2}{4}\le\frac{x^2+y^2}{2}\Rightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)

Khi do,ta co:

\(A\ge32-24\cdot1+9\cdot\frac{1}{2}=\frac{25}{2}\)

Dau bang xay ra khi va chi khi:\(x=y=\frac{1}{2}\)

P/S:E ko chac dau ah,e ms lm quen vs no thoi
 

\(VT\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\ge\frac{\left(4\left(x+y\right)+\frac{4}{x+y}-3\left(x+y\right)\right)^2}{2}\)

\(\ge\frac{\left(2.4-3.1\right)^2}{2}=\frac{25}{2}\)

đẳng thức xảy ra khi x = y = 1/2