cac ban tra loi di

có: \(\dfrac{1}{x^2+y^2}=\dfrac{1}{\left(x+y\right)^2-2xy}=\dfrac{1}{1-2xy}\)(1)

có \(\dfrac{1}{xy}=\dfrac{2}{2xy}\left(2\right)\)

từ(1)(2)=>A=\(\dfrac{1}{1-2xy}+\dfrac{2}{2xy}\ge\dfrac{\left(1+\sqrt{2}\right)^2}{1}=\left(1+\sqrt{2}\right)^2\)

=>Min A=(1+\(\sqrt{2}\))^2

b, ta có : \(x+y=1=>2x+2y=2\)

\(B=\dfrac{1}{x^2+y^2}+\dfrac{3}{4xy}=\dfrac{4}{4x^2+4y^2}+\dfrac{6}{8xy}\)\(\ge\dfrac{\left(2+\sqrt{6}\right)^2}{\left(2x+2y\right)^2}\)

\(=\dfrac{\left(2+\sqrt{6}\right)^2}{2^2}=\dfrac{5+2\sqrt{6}}{2}\)=>\(B\ge\dfrac{5+2\sqrt{6}}{2}\)

=>\(MinB=\dfrac{5+2\sqrt{6}}{2}\)

\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{4xy}+4xy+\dfrac{5}{4xy}\)

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

\(\ge\dfrac{4}{1^2}+2+\dfrac{5}{1^2}\) (do \(x+y\le1\))

\(=11\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)

Vậy GTNN của A là 11.

P/s  dùng bđt   1/a+1/b >=4/(a+b)

bạn chọn điểm rơi nha

tách 5/xy=1/2xy +...........

\(M=\frac{x^2+9y^2}{xy}-\frac{8y^2}{xy}\)

\(\ge\frac{2\sqrt{9x^2y^2}}{xy}-\frac{8.y.y}{xy}\)

\(\ge6-\frac{8.\frac{x}{3}.y}{xy}=6-\frac{8}{3}=\frac{10}{3}\)

Đẳng thức xảy ra khi x = 3y.

Vậy..

\(x\ge3y\Leftrightarrow\frac{x}{y}\ge3\)

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

\(\text{Đặt}\frac{x}{y}=a\Rightarrow a\ge3,M=a+\frac{1}{a}\)

Dùng điểm rơi a=3

\(M=\frac{8}{9}a+\frac{1}{9}a+\frac{1}{a}\ge\frac{8}{9}a+\frac{2}{3}\ge\frac{8}{3}+\frac{2}{3}=\frac{10}{3}\)

By Titu's Lemma we easy have:

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

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

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

\(=\frac{17}{4}\)

Mk xin b2 nha!

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

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

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

\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)

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

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

\(\Rightarrow\left(\sqrt{xy}+\frac{1}{4\sqrt{xy}}\right)^2+\frac{15}{16xy}+\frac{1}{2}\ge\frac{15}{16}\cdot4+\frac{1}{2}=\frac{17}{4}\)

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