a) Ta có : \(A=\left|x+1\right|+\left|y-2\right|\)

\(\ge\left|x+1+y-2\right|\)

\(=\left|x+y-1\right|=\left|5-1\right|=\left|4\right|=4\)

Dấu "=" xảy ra <=> (x + 1)(y - 2) \(\ge\)0

Vậy Min A = 4 <=>  (x + 1)(y - 2) \(\ge\)0

Bạn có thể làm cách sau tuy hơi dài

Lấy x=1-y thay vào P rồi phá ngoặc lẫn dấu ra.

Ta sẽ tìm được GTNN của P.

Bài này hoàn toàn có thể giải bằng BĐT Cổ điển.

BĐT Cauchy-schwarz( Bunhiacopxki):

\(P\ge\frac{1}{2}.\left(2x+2y+\frac{1}{x}+\frac{1}{y}\right)^2\)

Việc còn lại không khó! :)

Ghi thêm điều kiện x,y dương nhé

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

P>= 2.\(\left\{4+\frac{2}{x}+\frac{2}{y}+\frac{1}{xy}\right\}^{ }\)

P>=8 + 4\(\left\{\frac{1}{x}+\frac{1}{y}\right\}^{ }\) + \(\frac{2}{xy}\)

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

P>= 8+ \(\frac{4}{xy}\)+\(\frac{2}{xy}\)

P>= 8+ \(\frac{6}{xy}\)>= 8+ 6.\(\frac{4}{\left(x+y\right)^2}\)>= 8 + 6.4= 32

dấu = xảy ra khi x=y =\(\frac{1}{2}\)

\(P=\left(2x+\dfrac{1}{x}\right)^2+9+\left(2y+\dfrac{1}{y}\right)^2+9-18\)

\(P\ge2\sqrt{9\left(2x+\dfrac{1}{x}\right)^2}+2\sqrt{9\left(2y+\dfrac{1}{y}\right)^2}-18\)

\(P\ge12x+12y+\dfrac{6}{x}+\dfrac{6}{y}-18\)

\(P\ge6\left(4x+\dfrac{1}{x}\right)+6\left(4y+\dfrac{1}{y}\right)-12\left(x+y\right)-18\)

\(P\ge6.2\sqrt{\dfrac{4x}{x}}+6.2\sqrt{\dfrac{4y}{y}}-12.1-18=18\)

\(P_{min}=18\) khi \(x=y=\dfrac{1}{2}\)

-Áp dụng BĐT AM-GM ta có:

\(xy\le\dfrac{\left(x+y\right)^2}{4}\Leftrightarrow xy\le\dfrac{2^2}{4}=1\)

\(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}=\dfrac{2^2}{2}=2\)

\(A=\left(2x+\dfrac{1}{x}\right)^2+\left(2y+\dfrac{1}{y}\right)^2+2001=4x^2+4+\dfrac{1}{x^2}+4y^2+4+\dfrac{1}{y^2}+2001=4\left(x^2+y^2\right)+\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+2009\ge4.2+2.\dfrac{1}{xy}+2009\ge8+2.\dfrac{1}{1}+2009=2019\)

\(A=2019\Leftrightarrow x=y=1\)

-Vậy \(A_{min}=2019\)

Áp dụng Bunyakovsky, ta có :

\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x.1+y.1\right)^2=1\)

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

=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)

Mấy cái kia tương tự 

a, \(A=\left|x+1\right|+\left|y-2\right|\)

\(A=\left|x+1\right|+\left|5-x-2\right|\)

\(A=\left|x+1\right|+\left|3-x\right|\ge x+1+3-x=4\)

Dấu " = " sảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\3-x\ge0\end{matrix}\right.\Leftrightarrow-1\le x\le3\)