\(S=\frac{1}{x^2+y^2}+\frac{2}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{3}{2xy}+4xy\)

\(\ge\frac{\left(1+1\right)^2}{\left(x+y\right)^2}+\frac{3}{2xy}+4xy\ge\frac{4}{\frac{1}{4}}+\frac{3}{2xy}+384xy-380xy\)

\(\ge16+2\cdot24-380xy=64-380xy\)

+) \(\frac{1}{2}\ge x+y\ge2\sqrt{xy}\Rightarrow\frac{1}{4}\ge4xy\Leftrightarrow\frac{1}{16}\ge xy\)

\(\Rightarrow-380xy\ge380\cdot\frac{1}{16}=23.75\)

\(\Rightarrow S\ge64-23.75=40.25\)

Dấu = xảy ra khi x=y=1/4

Tại sao \(\frac{1}{x^2+y^2}+\frac{1}{2xy}\le\frac{\left(1+1\right)^2}{\left(x+y\right)^2}\)  ?

Bạn xem BDT Cauchy-Schwarz nhé!

Ta có : \(P=\frac{1}{x^2+y^2}+\frac{2}{xy}=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{3}{2xy}\)

Áp dụng bđt \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)được :\(\frac{1}{x^2+y^2}+\frac{1}{2xy}\ge\frac{4}{\left(x+y\right)^2}\ge4\)

Áp dụng bđt \(\frac{1}{ab}\ge\frac{4}{\left(a+b\right)^2}\)được : \(\frac{3}{2xy}\ge\frac{3}{2}.\frac{4}{\left(x+y\right)^2}\ge6\)

Suy ra \(P\ge10\)

Dấu "=" xảy ra khi và chỉ khi \(\hept{\begin{cases}x+y=1\\x=y\end{cases}}\)\(\Leftrightarrow x=y=\frac{1}{2}\)

Vậy Min P = 10 khi x = y = 1/2

Suy ra P≥10

Dấu "=" xảy ra khi và chỉ khi {

⇔x=y=12 

Vậy Min P = 10 khi x = y = 1/2

Từ BĐT \(\left(x+y\right)^2\ge4xy\) ta suy ra \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) và \(\frac{1}{xy}\ge\frac{4}{\left(x+y\right)^2}\)

Ta có : \(P=\frac{20}{x^2+y^2}+\frac{11}{xy}=20\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1}{xy}\ge20.\frac{4}{\left(x+y\right)^2}+\frac{4}{\left(x+y\right)^2}\ge\frac{80}{4}+\frac{4}{4}=21\)

Dấu "=" xảy ra khi x = y = 1

Vậy Min P = 21 khi x = y = 1

Ta có :

\(P=\frac{20}{x^2+y^2}+\frac{11}{xy}\)

\(=20.\left[\frac{1}{x^2+y^2}+\frac{1}{2xy}\right]+\frac{1}{xy}\)

\(\ge20\cdot\frac{4}{x^2+y^2+2xy}+\frac{4}{\left(x+y\right)^2}\)

\(\ge20\cdot\frac{4}{2^2}+\frac{4}{2^2}=21\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=1\)

Vậy \(P_{min}=21\) khi \(x=y=1\)

Ta có:

\(P=20\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1}{xy}\)

\(\ge20\cdot\frac{4}{\left(x+y\right)^2}+\frac{4}{\left(x+y\right)^2}\ge21\)

\(\Rightarrow P\ge21\)

Dấu = khi x=y=1

1. \(1=x^2+y^2\ge2xy\Rightarrow xy\le\frac{1}{2}\)

 \(A=-2+\frac{2}{1+xy}\ge-2+\frac{2}{1+\frac{1}{2}}=-\frac{2}{3}\)

max A = -2/3 khi x=y=\(\frac{\sqrt{2}}{2}\)

\(\frac{1}{xy}+\frac{1}{xz}=\frac{1}{x}\left(\frac{1}{y}+\frac{1}{z}\right)\ge\frac{1}{x}.\frac{4}{y+z}=\frac{4}{\left(4-t\right)t}=\frac{4}{4-\left(t-2\right)^2}\ge1\) với t = y+z => x =4 -t

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

\(\sqrt{A}=\sqrt{x^2+y^2}\le\sqrt{x^2}+\sqrt{y^2}=x+y=1\)( \(\sqrt{a+b}\le\sqrt{a}+\sqrt{b}\))

=> A\(\le1\) => Max A = 1 khi x =0;y =1 hoặc x =1 ; y =0

1/a/
\(A=\frac{2}{xy}+\frac{3}{x^2+y^2}=\left(\frac{1}{xy}+\frac{1}{xy}+\frac{4}{x^2+y^2}\right)-\frac{1}{x^2+y^2}\)

\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}-\frac{1}{\frac{\left(x+y\right)^2}{2}}=16-2=14\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

b/

\(4B=\frac{4}{x^2+y^2}+\frac{8}{xy}+16xy=\left(\frac{4}{x^2+y^2}+\frac{1}{xy}+\frac{1}{xy}\right)+\left(\frac{1}{xy}+16xy\right)+\frac{5}{xy}\)

\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}+2\sqrt{\frac{1}{xy}.16xy}+\frac{5}{\frac{\left(x+y\right)^2}{4}}\)

\(=16+8+20=44\)

\(\Rightarrow B\ge11\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

2/

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

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

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

Áp dụng BĐT Cauchy-schwarz ta có:

\(8\ge x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)

\(\Leftrightarrow x+y\ge4\)

Dấu " = " xảy ra <=> x=y=2

Áp dụng BĐT Cauchy-schwarz ta có:

\(A\ge\frac{4}{x+y}\ge\frac{4}{4}=1\)

Dấu " = " xảy ra <=> x=y=2

Hình như anh kudo shinichi ngược dấu một xíu thì phải ạ: \(8\ge\frac{\left(x+y\right)^2}{2}\Rightarrow\left(x+y\right)\le4\) chứ ạ?Dẫn đến 

khúc sau ngược dấu.Nếu em sai thì xin thông ảm cho ạ. Lời giải của em đây:

\(A\ge\frac{4}{x+y}=\frac{16}{4x+4y}\ge\frac{16}{x^2+4+y^2+4}\) (BĐT Cô si hay AM-GM gì đó: \(x^2+4\ge2\sqrt{x^2.4}=2.2.x=4x;...\))

\(=\frac{16}{8+8}=1\).Dấu "=" xảy ra khi x = y = 2.

Vậy min A = 1 khi x =y = 2

tth: Chỗ đó a gõ nhầm nhưng khúc sau không ngược dấu e nhé

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

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

Dấu "=" <=> x= y = 1/2

\(2,A=\frac{x^2+y^2}{xy}=\frac{x}{y}+\frac{y}{x}=\left(\frac{x}{9y}+\frac{y}{x}\right)+\frac{8x}{9y}\ge2\sqrt{\frac{x}{9y}.\frac{y}{x}}+\frac{8.3y}{9y}\)

                                                                                                  \(=2\sqrt{\frac{1}{9}}+\frac{8.3}{9}=\frac{10}{3}\)

Dấu "=" <=> x = 3y

bài 3 min hay max ?