\(A=\frac{4}{4x^2+9y^2}+\frac{9}{xy}=\frac{4}{4x^2+9y^2}+\frac{54}{6xy}\)

Đặt \(\left\{{}\begin{matrix}2x=a\\3y=b\end{matrix}\right.\Rightarrow A=\frac{4}{a^2+b^2}+\frac{54}{ab}\)

\(A=\frac{4}{a^2+b^2}+\frac{4}{2ab}+\frac{52}{ab}\)

\(A=4\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\frac{52}{ab}\)

\(\ge\frac{16}{\left(a+b\right)^2}+\frac{52}{\frac{\left(a+b\right)^2}{4}}\ge4+52=56\)

\("="\Leftrightarrow a=b\Leftrightarrow2x=3y\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{3}\end{matrix}\right.\)

\(A=\frac{4}{4x^2+9y^2}+\frac{4}{12xy}+\frac{52}{2x.3y}\ge\frac{16}{4x^2+9y^2+12xy}+\frac{52}{\frac{\left(2x+3y\right)^2}{4}}\)

\(A\ge\frac{16}{\left(2x+3y\right)^2}+\frac{208}{\left(2x+3y\right)^2}\ge\frac{16}{4}+\frac{208}{4}=56\)

\(\Rightarrow A_{min}=56\) khi \(\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{3}\end{matrix}\right.\)

Áp dụng bất dẳng thức AM-GM ta có:

\(A=\frac{4}{4x^2+9y^2}+\frac{4}{12xy}+\frac{208}{24xy}\ge\frac{\left(2+2\right)^2}{4x^2+9y^2+12xy}+\frac{208}{\left(2x+3y\right)^2}=\frac{16}{\left(2x+3y\right)^2}+\frac{208}{\left(2x+3y\right)^2}\ge\frac{16}{4}+\frac{208}{4}=56\)

Đẳng thức xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}2x=3y\\2x+3y=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{3}\end{matrix}\right.\)

Vậy A_min = 56 \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{3}\end{matrix}\right.\)

Theo cô-si thì \(2\sqrt{2x.3y}\le2x+3y\le2\Rightarrow xy\le\frac{1}{6}\)

\(A=\frac{4}{4x^2+9y^2}+\frac{9}{xy}=\frac{4}{4x^2+9y^2}+\frac{4}{12xy}+\frac{26}{3xy}\)

                                            \(\ge\frac{\left(2+2\right)^2}{4x^2+9y^2+12xy}+\frac{26}{\frac{3.1}{6}}\)

                                            \(=\frac{14}{\left(2x+3y\right)^2}+\frac{26.6}{3}=56\)

\("="\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{3}\end{cases}}\)

ta thấy \(A=\frac{4}{4x^2+9y^2}+\frac{9}{xy}=\frac{4}{4x^2+9y^2}+\frac{4}{12xy}+\frac{26}{3xy}\ge\frac{16}{\left(2x+3y\right)^2}+\frac{26}{3xy}\)(1)

lại có \(2x+3y\le2\Leftrightarrow\left(2x+3y\right)^2\le4\Leftrightarrow4x^2+9y^2+12xy\le4\left(2\right)\)

mặt khác \(4x^2+9y^2\ge12xy\)(theo Bất Đẳng Thức Cosi cho x,y>0) (3)

từ (1) và (2) => \(12xy+12xy\le4\Leftrightarrow3xy\le\frac{1}{2}\left(4\right)\)

từ (1) và (4) => \(A\ge\frac{16}{4}+\frac{26}{\frac{1}{2}}=4+52=56\)

dấu "=" xảy ra khi \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{3}\end{cases}}\)

Đặt \(\hept{\begin{cases}2x=a\left(a>0\right)\\3y=b\left(b>0\right)\end{cases}}\)

\(\Rightarrow2x+3y=a+b\le2,x.y=\frac{ab}{6}\)

\(\Rightarrow P=\frac{4}{a^2+b^2}+\frac{9}{\frac{ab}{6}}=\frac{4}{a^2+b^2}\ne\frac{54}{ab}\)

Vì \(a>0,b>0\)

Nên áp dụng BĐT cô-si ta có:\(a+b\ge2\sqrt{ab}\)

Mà \(a+b\le2\Rightarrow2\sqrt{ab}\le2\Rightarrow\sqrt{ab}\le1\Rightarrow ab\le1\)

Áp dụng BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)với x > 0 , y > 0 

\(\Rightarrow\frac{1}{a^2+b^2}+\frac{1}{2ab}\ge\frac{4}{a^2+b^2+2ab}=\frac{4}{\left(a+b\right)^2}\ge1\)

\(\Rightarrow\frac{4}{a^2+b^2}+\frac{4}{2ab}\ge4\)

\(\Rightarrow P=\frac{4}{a^2+b^2}+\frac{4}{2ab}+\frac{52}{ab}\)

\(P\ge4+52=56\)

\(\Rightarrow MinP=56\Leftrightarrow\hept{\begin{cases}a=b\\a+b=2\\a.b=1\end{cases}}\Leftrightarrow\hept{a=b=1\Leftrightarrow2x=3y=1\Leftrightarrow x=\frac{1}{2},y=\frac{1}{3}}\)

\(A=\frac{4}{4x^2+9y^2}+\frac{4}{12xy}+\frac{52}{2x.3y}\)

\(A\ge\frac{16}{4x^2+9y^2+12xy}+\frac{52.4}{\left(2x+3y\right)^2}=\frac{224}{\left(2x+3y\right)^2}\ge\frac{224}{4}=56\)

\(A_{min}=56\) khi \(\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{3}\end{matrix}\right.\)

\(A=\frac{4}{4x^2+9y^2}+\frac{4}{12xy}+\frac{52}{2x.3y}\)

\(A\ge\frac{16}{4x^2+9y^2+12xy}+\frac{52.4}{\left(2x+3y\right)^2}\)

\(A\ge\frac{224}{\left(2x+3y\right)^2}\ge\frac{224}{4}=56\)

\(A_{min}=56\) khi \(\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{3}\end{matrix}\right.\)

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

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

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 ?