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 \(MinM=20\)

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.

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

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 \(\Leftrightarrow x=y=\dfrac{1}{2}\)

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 \(x=y=\dfrac{1}{2}\)

\(y=x+\dfrac{1}{x}-5\ge2\sqrt{\dfrac{x}{x}}-5=-3\)

\(y_{min}=-3\) khi \(x=1\)

\(y=4x^2+\dfrac{1}{2x}+\dfrac{1}{2x}-4\ge3\sqrt[3]{\dfrac{4x^2}{2x.2x}}-4=-1\)

\(y_{min}=-1\) khi \(x=\dfrac{1}{2}\)

\(y=x+\dfrac{4}{x}\Rightarrow y'=1-\dfrac{4}{x^2}=0\Rightarrow x=-2\)

\(y\left(-2\right)=-4\Rightarrow\max\limits_{x>0}y=-4\) khi \(x=-2\)

\(\dfrac{\left(x+y+1\right)^2}{xy+x+y}\ge\dfrac{3\left(xy+x+y\right)}{xy+x+y}=3\)

\(\Rightarrow A=\dfrac{8\left(x+y+1\right)^2}{9\left(xy+x+y\right)}+\dfrac{\left(x+y+1\right)^2}{9\left(xy+x+y\right)}+\dfrac{xy+x+y}{\left(x+y+1\right)^2}\)

\(A\ge\dfrac{8}{9}.3+2\sqrt{\dfrac{\left(x+y+1\right)^2\left(xy+x+y\right)}{\left(xy+x+y\right)\left(x+y+1\right)^2}}=\dfrac{10}{3}\)

Dấu "=" xảy ra khi \(x=y=1\)

Ta có: \(xy\le\dfrac{1}{4}\left(x+y\right)^2=\dfrac{1}{4}\times1^2=\dfrac{1}{4}\)

\(\Rightarrow x^2y^2\le\dfrac{1}{16}\)

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

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

\(\ge\dfrac{1}{16}+1+1+\dfrac{1}{\dfrac{1}{16}}=\dfrac{289}{16}\)

Dấu "=" xảy ra <=> x = y = 0,5

Vậy Min A = 18,0625 <=> x = y = 0,5

mình khẳng định cách làm này chắc chắn đúng

A=(x² +1/y²)(y² +1/x²)=(xy)²+\(\dfrac{1}{xy^2}\)+2

ta có x+y=1 mà x+y \(\ge\)2\(\sqrt{xy}\)nên 1 \(\ge\)2\(\sqrt{xy}\)

nên 1/2 \(\ge\)\(\sqrt{xy}\) =>1/4\(\ge\)xy=>\(\dfrac{1}{16}\)\(\ge\)(xy)²

sau đó ta sử dụng phương pháp chọn điểm rơi để thêm bớt cho phù hợp.

ta thấy gtnn xảy ra <=>x=y=1/2 hay (xy)²=1/16

để bảo toàn cho giá trị nhỏ nhất xảy ra với điều kiện đè bài đã cho là x+y=1 thì ta đặt hằng số \(\alpha\)sao cho:

đặt \(\dfrac{\alpha}{xy^2}\)=xy²

cho xy²=\(\dfrac{1}{16}\)thì\(\alpha\)=\(\dfrac{1}{256}\)

ta có lời giải A=(\(\dfrac{1}{xy^2}\)-\(\dfrac{\dfrac{1}{256}}{xy^2}\))+(\(\dfrac{\dfrac{1}{256}}{xy^2}\)+xy²)+2

áp dụng bất đẳng thức cosy a²+b²\(\ge\)2ab ta có

\(\dfrac{\dfrac{1}{256}}{xy^2}\)+xy²\(\ge\)2\(\dfrac{\dfrac{1}{16}}{xy}\).xy=\(\dfrac{1}{8}\)

ta đã chứng minh \(\dfrac{1}{16}\)\(\ge\)xy² nên ta có

\(\dfrac{1}{xy^2}\)-\(\dfrac{\dfrac{1}{256}}{xy^2}\)=\(\dfrac{\dfrac{255}{256}}{xy2}\)\(\ge\)\(\dfrac{\dfrac{255}{256}}{\dfrac{1}{16}}\)=\(\dfrac{255}{16}\)

nên A\(\ge\)\(\dfrac{1}{8}\)+\(\dfrac{255}{16}\)+2=\(\dfrac{289}{16}\)

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

vậy min A=\(\dfrac{289}{16}\)tại x=y=\(\dfrac{1}{2}\)

Phương An

Vấn đề thiếu sót

Vì Sao ? khi xy =16 t lại kết luận được A nhỏ nhất --> ép buộc đang cơ sở lý luận toán học