Ta có \(\left(x^2+\dfrac{1}{y^2}\right)\left(y^2+\dfrac{1}{x^2}\right)=x^2y^2+1+1+\dfrac{1}{x^2y^2}=x^2y^2+2+\dfrac{1}{x^2y^2}=\dfrac{x^4y^4+2x^2y^2+1}{x^2y^2}=\dfrac{\left(x^2y^2+1\right)^2}{\left(xy\right)^2}=\left(\dfrac{x^2y^2+1}{xy}\right)^2=\left(xy+\dfrac{1}{xy}\right)^2=\left(xy+\dfrac{1}{16xy}+\dfrac{15}{16xy}\right)^2\)

Áp dụng bđt cosi, ta có \(xy+\dfrac{1}{16xy}\ge2\sqrt{xy.\dfrac{1}{16xy}}=2\sqrt{\dfrac{1}{16}}=2.\dfrac{1}{4}=\dfrac{1}{2}\)

\(2\sqrt{xy}\le\left(x+y\right)^2\Leftrightarrow\sqrt{xy}\le\dfrac{\left(x+y\right)^2}{2}=\dfrac{1}{2}\Leftrightarrow xy\le\dfrac{1}{4}\Leftrightarrow\dfrac{15}{16xy}\ge\dfrac{15}{4}\)

Vậy \(xy+\dfrac{1}{16xy}+\dfrac{15}{16xy}\ge\dfrac{1}{2}+\dfrac{15}{4}=\dfrac{17}{4}\Leftrightarrow\left(xy+\dfrac{1}{16xy}+\dfrac{15}{16xy}\right)^2\ge\dfrac{289}{16}\)

Dấu bằng xảy ra khi \(\left\{{}\begin{matrix}x+y=1\\xy=\dfrac{1}{16xy}\\x=y\end{matrix}\right.\)\(\Leftrightarrow\)\(x=y=0,5\)

Vậy GTNN của \(\left(x^2+\dfrac{1}{y^2}\right)\left(y^2+\dfrac{1}{x^2}\right)\)=\(\dfrac{289}{16}\) và xảy ra khi x=y=0,5

Ta có \(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left(\dfrac{\left(a+b\right)^2}{2}\right)^2}{2}=\dfrac{\left(a+b\right)^4}{8}\). Áp dụng cho biểu thức A, suy ra \(A\ge\dfrac{\left(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\right)^4}{8}\). Ta tìm GTNN của \(P=x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\). Ta có 

\(P=x^2+\dfrac{1}{16x^2}+y^2+\dfrac{1}{16y^2}+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+2\)

\(P\ge2\sqrt{x^2.\dfrac{1}{16x^2}}+2\sqrt{y^2.\dfrac{1}{16y^2}}+\dfrac{15}{16}\left(\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2}{2}\right)+2\)

    \(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{15}{16}.\left(\dfrac{4^2}{2}\right)+2\) \(=\dfrac{21}{2}\). Do đó \(P\ge\dfrac{21}{2}\) \(\Leftrightarrow A\ge\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\). Vậy GTNN của A là \(\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\), ĐTXR \(\Leftrightarrow x=y=\dfrac{1}{2}\)

Gợi ý: \(\dfrac{a^4+b^4}{2}\ge\left(\dfrac{a+b}{2}\right)^4\)

Ta có \(B\ge\dfrac{\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)^2}{2}\) \(=\dfrac{\left(1+\dfrac{1}{xy}\right)^2}{2}\)

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

\(\Rightarrow B\ge\dfrac{\left(1+4\right)^2}{2}=\dfrac{25}{2}\)

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

Vậy GTNN của B là \(\dfrac{25}{2}\) khi \(x=y=\dfrac{1}{2}\)

C₁:

\(x,y>0\)

\(M=\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2=x^2+2+\dfrac{1}{x^2}+y^2+2+\dfrac{1}{y^2}=\left(x^2+\dfrac{1}{16x^2}\right)+\left(y^2+\dfrac{1}{16y^2}\right)+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+4\)Theo BĐT AM-GM (Caushy) ta có:

\(M=\left(x^2+\dfrac{1}{16x^2}\right)+\left(y^2+\dfrac{1}{16y^2}\right)+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+4\ge2\sqrt{x^2.\dfrac{1}{16x^2}}+2\sqrt{y^2.\dfrac{1}{16y^2}}+\dfrac{15}{16}.2\sqrt{\dfrac{1}{x^2}.\dfrac{1}{y^2}}+4=\dfrac{1}{2}+\dfrac{1}{2}+4+\dfrac{15}{4}.\dfrac{1}{xy}\ge5+\dfrac{15}{4}.\dfrac{1}{\left(\dfrac{x+y}{2}\right)^2}\ge5+\dfrac{15}{4}.\dfrac{1}{\left(\dfrac{1}{2}\right)^2}=20\)Đẳng thức xảy ra \(\left\{{}\begin{matrix}x^2=\dfrac{1}{16}x^2\\y^2=\dfrac{1}{16}y^2\\x+y=1\\x,y>0\end{matrix}\right.\Leftrightarrow x=y=\dfrac{1}{2}\)

Vậy \(MinM=20\)

Lời giải:

Ta có: \(A=\left(1-\frac{4}{x^2}\right)\left(1-\frac{4}{y^2}\right)\)

\(A=\frac{(x^2-4)(y^2-4)}{x^2y^2}\)

\(A=\frac{[x^2-(x+y)^2][y^2-(x+y)^2]}{x^2y^2}=\frac{(-y)(2x+y)(-x)(2y+x)}{x^2y^2}\)

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

\(A=5+\frac{2(x^2+y^2)}{xy}=5+\frac{2(x-y)^2+4xy}{xy}=9+\frac{2(x-y)^2}{xy}\)

Thấy rằng \(x,y>0; (x-y)^2\geq 0\Rightarrow \frac{2(x-y)^2}{xy}\geq 0\)

\(\Rightarrow A\geq 9\) hay \(A_{\min}=9\)

Dấu bằng xảy ra khi \(x=y=1\)

\(A=\left(1-\dfrac{4}{x^2}\right)\left(1-\dfrac{4}{y^2}\right)=\left(1-\dfrac{2}{x}\right)\left(1+\dfrac{2}{x}\right)\left(1-\dfrac{2}{y}\right)\left(1+\dfrac{2}{y}\right)\)

\(A=\left(\dfrac{x-2}{x}\right)\left(\dfrac{x+2}{x}\right)\left(\dfrac{y-2}{y}\right)\left(\dfrac{y+2}{x}\right)\)

x+y=2 => x-2 =-y; y-2=-x

\(A=\left(\dfrac{-y}{y}\right)\left(\dfrac{x+2}{x}\right)\left(\dfrac{-x}{x}\right)\left(\dfrac{y+2}{y}\right)=\left(\dfrac{x+2}{x}\right)\left(\dfrac{y+2}{y}\right)\)

\(A=\dfrac{xy+2\left(x+y\right)+4}{xy}=\dfrac{xy+8}{xy}=1+\dfrac{8}{xy}\)

áp dụng co si cho 2 số dương x;y

\(2=\left(x+y\right)\ge2\sqrt{xy}\Rightarrow xy\le1\Rightarrow\dfrac{1}{xy}\ge1\)

\(A\ge1+8.1=9\)đẳng thức x=y=1

Lời giải:

Áp dụng BĐT AM-GM:
$M\geq 2\sqrt{\frac{1}{xy}}.\sqrt{1+x^2y^2}=2\sqrt{\frac{x^2y^2+1}{xy}}$
$=2\sqrt{xy+\frac{1}{xy}}$

Áp dụng BĐT AM-GM tiếp:

$1\geq x+y\geq 2\sqrt{xy}\Rightarrow xy\leq \frac{1}{4}$
$xy+\frac{1}{xy}=(xy+\frac{1}{16xy})+\frac{15}{16xy}$

$\geq 2\sqrt{xy.\frac{1}{16xy}}+\frac{15}{16xy}$

$\geq 2\sqrt{\frac{1}{16}}+\frac{15}{16.\frac{1}{4}}=\frac{17}{4}$

$\Rightarrow M\geq 2\sqrt{\frac{17}{4}}=\sqrt{17}$

Vậy $M_{\min}=\sqrt{17}$. Giá trị này đạt tại $x=y=\frac{1}{2}$

Sửa đề:

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

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

\(=8+4\left(x^2+y^2\right)+\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)

\(\ge8+4.\dfrac{\left(x+y\right)^2}{2}+\dfrac{2}{xy}\)

\(\ge8+4.\dfrac{\left(x+y\right)^2}{2}+\dfrac{2}{\dfrac{\left(x+y\right)^2}{4}}\)

\(=8+4.\dfrac{1}{2}+\dfrac{2}{\dfrac{1}{4}}=18\)

Vậy GTNN là P = 18 đạt được khi \(x=y=\dfrac{1}{2}\)

Hình như đầu bài sai hay sao ý đáng ra phải là

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

Mình giải theo đầu bài chữa nhé:

Vì x, y > 0

Áp dụng BĐT-Cauchy, ta có:

\(2x+\dfrac{1}{x}\ge\sqrt{2x+\dfrac{1}{x}}\)

\(2y+\dfrac{1}{y}\ge\sqrt{2y.\dfrac{1}{y}}\)

\(\Rightarrow P\ge\left(\sqrt{2x+\dfrac{1}{x}}\right)^2+\left(\sqrt{2y+\dfrac{1}{y}}\right)^2\)

\(P\ge\left(2\sqrt{2}\right)^2+\left(2\sqrt{2}\right)^2\)

\(P\ge8+8=16\)

GTNN của P là 16 khi \(x=y=\dfrac{1}{2}\)

Áp dụng bất đẳng thức AM - GM:

\(P\ge3\sqrt[3]{\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}\).

Áp dụng bất đẳng thức AM - GM ta có:

\(xy+1=xy+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\ge5\sqrt[5]{\dfrac{xy}{4^4}}\).

Tương tự: \(yz+1\ge5\sqrt[5]{\dfrac{yz}{4^4}};zx+1\ge5\sqrt[5]{\dfrac{zx}{4^4}}\).

Do đó \(\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)\ge125\sqrt[5]{\dfrac{\left(xyz\right)^2}{4^{12}}}\)

\(\Rightarrow\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{1}{4^{12}\left(xyz\right)^3}}\).

Mà \(xyz\le\dfrac{\left(x+y+z\right)^3}{27}=\dfrac{1}{8}\)

Nên \(\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{8^3}{4^{12}}}=125\sqrt[5]{\dfrac{1}{2^{15}}}=\dfrac{125}{8}\)

\(\Rightarrow P\ge\dfrac{15}{2}\).

Vậy...

Áp dụng bất đẳng thức AM - GM:

P≥33√(xy+1)(yz+1)(zx+1)xyz.

Áp dụng bất đẳng thức AM - GM ta có:

xy+1=xy+14+14+14+14≥55√xy44.

Tương tự: yz+1≥55√yz44;zx+1≥55√zx44.

Do đó (xy+1)(yz+1)(zx+1)≥1255√(xyz)2412

⇒(xy+1)(yz+1)(zx+1)xyz≥1255√1412(xyz)3.

Mà xyz≤(x+y+z)327=18

Nên  (xy+1)(yz+1)(zx+1)xyz≥1255√83412=1255√1215=1258 

⇒P≥152.