\(A=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1001}{2xy}\ge\frac{4}{x^2+y^2+2xy}+\frac{1001}{\frac{2\left(x+y\right)^2}{4}}\)

\(A\ge\frac{4}{\left(x+y\right)^2}+\frac{2002}{\left(x+y\right)^2}=\frac{2006}{\left(x+y\right)^2}\ge2006\)

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

\(A=\frac{1}{x^2+y^2}+\frac{501}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+\frac{500}{xy}\)

\(\ge\frac{5}{\left(x+y\right)^2}+\frac{500}{\frac{\left(x+y\right)^2}{2}}=5+1000=1005\)

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

đoán là sai

\(A=\frac{1}{x^2+y^2}+\frac{501}{xy}\)

\(=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1001}{2xy}\ge\frac{4}{\left(x+y\right)^2}+\frac{1001}{\frac{\left(x+y\right)^2}{2}}\ge4+2002=2006\)

Dấu "=" xảy ra khi  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

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 <=> \(x=y=\frac{1}{2}\)

Ta có: \(1\ge x+y\ge2\sqrt{xy}\Rightarrow1\ge4xy\Rightarrow\frac{1}{xy}\ge4\)

\(\Rightarrow P\ge2\sqrt{\frac{1}{xy}}\cdot\sqrt{1+x^2y^2}=2\sqrt{\frac{1}{xy}+xy}\)

Mà \(\frac{1}{xy}+xy=\frac{15}{16}\cdot\frac{1}{xy}+\frac{1}{16xy}+xy\)

\(\ge\frac{15}{16}\cdot4+2\sqrt{\frac{1}{16xy}\cdot xy}=\frac{15}{16}\cdot4+\frac{2}{4}=\frac{17}{4}\)

\(\Rightarrow P\ge2\cdot\frac{\sqrt{17}}{2}=\sqrt{17}\) xảy ra khi \(x=y=\frac{1}{2}\)

v~ máy mk ko gõ dc chữ "x" 

cám ơn thắng nhìu

Với \(x+y< 1\) thì biểu thức ko tồn tại min

Nó chỉ tồn tại min khi \(x+y\le1\)

1 thách dám tích

\(A=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\ge\frac{4}{\left(x+y\right)^2}+\frac{1}{2xy}\\ =\frac{1}{4}+\frac{1}{2xy}\ge\frac{1}{4}+\frac{1}{8}=\frac{3}{8}\)

Dấu = xảy ra khi x=y=2

vay la ??????????????????????????