Xét: \(\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}-\frac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}=\frac{x^4-y^4}{\left(x^2+y^2\right)\left(x+y\right)}\)\(=\frac{\left(x^2+y^2\right)\left(x^2-y^2\right)}{\left(x^2+y^2\right)\left(x+y\right)}=\frac{\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)}{\left(x^2+y^2\right)\left(x+y\right)}=x-y\)(1)

Tương tự, ta có: \(\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}-\frac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}=y-z\)(2); \(\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}-\frac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}=z-x\)(3)

Cộng theo vế của 3 đẳng thức (1), (2), (3), ta được:

\(\left[\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}\right]\)\(-\left[\frac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\right]=0\)

\(\Rightarrow\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}\)\(=\frac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)

Mà \(A=\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}\)nên \(2A=\frac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4+z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4+x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)\(\ge\frac{\frac{\left(y^2+z^2\right)^2}{2}}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{\frac{\left(y^2+z^2\right)^2}{2}}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{\frac{\left(z^2+x^2\right)^2}{2}}{\left(z^2+x^2\right)\left(z+x\right)}\)

\(=\frac{1}{2}\left(\frac{x^2+y^2}{x+y}+\frac{y^2+z^2}{y+z}+\frac{z^2+x^2}{z+x}\right)\)\(\ge\frac{1}{2}\left(\frac{\frac{\left(x+y\right)^2}{2}}{x+y}+\frac{\frac{\left(y+z\right)^2}{2}}{y+z}+\frac{\frac{\left(z+x\right)^2}{2}}{z+x}\right)\)\(=\frac{1}{4}\left[\left(x+y\right)+\left(y+z\right)+\left(z+x\right)\right]=\frac{1}{2}\left(x+y+z\right)=\frac{1}{2}\)(Do theo giả thiết thì x + y + z = 1)

\(\Rightarrow A\ge\frac{1}{4}\)

Đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\)

Bài này t làm rồi, "nhẹ" không ấy mà :|

Dự đoán khi \(x=y=z=\frac{1}{3}\Rightarrow A=\frac{1}{4}\). Ta c/m nó là GTNN của A

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(A=Σ\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{Σ\left(x^2+y^2\right)\left(x+y\right)}\)

Cần chứng minh BĐT \(\frac{\left(x^2+y^2+z^2\right)^2}{Σ\left(x^2+y^2\right)\left(x+y\right)}\ge\frac{x+y+z}{4}\)

\(\Leftrightarrow4\left(x^2+y^2+z^2\right)^2\ge\left(x+y+z\right)Σ\left(2x^3+x^2y+x^2z\right)\)

\(\LeftrightarrowΣ\left(2x^4-3x^3y-3x^3z+6x^2y^2-2x^2yz\right)\ge0\)

\(\LeftrightarrowΣ\left(2x^4-3x^3y-3x^3z+4x^2y^2\right)+Σ\left(2x^2y^2-2x^2yz\right)\ge0\)

\(\LeftrightarrowΣ\left(x^4-3x^3y+4x^2y^2-3xy^3+y^4\right)+Σ\left(x^2z^2-2z^2xy+y^2z^2\right)\ge0\)

\(\LeftrightarrowΣ\left(x-y\right)^2\left(x^2-xy+y^2\right)+Σz^2\left(x-y\right)^2\ge0\)

BĐT cuối đúng tức ta có \(A_{Min}=\frac{1}{4}\Leftrightarrow x=y=z=\frac{1}{3}\)

P/s: Nguồn lời giải Câu hỏi của Vo Trong Duy - Toán lớp 9 - Học toán với OnlineMath, rảnh quá ngồi gõ lại :V

\(VT=3\left(9x^2-12x+4\right)+\frac{8x}{1-x}=27x^2-36x+12+\frac{8x}{1-x}\)

\(=27x^2-36x+4+\frac{8}{1-x}=27x^2-18x-6+8\left(1-x\right)+\frac{8}{1-x}\)

\(=27x^2-18x+3+8\left(1-x\right)+\frac{8}{1-x}-9\)

\(=3\left(3x-1\right)^2+8\left(1-x\right)+\frac{8}{1-x}-9\)

\(\Rightarrow VT\ge2\sqrt{8^2}-9=7\)

Dấu " = " xảy ra khi \(x=\frac{1}{3}\)

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

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

Cô-si : \(1=x+y\ge2\sqrt{xy}\Leftrightarrow xy\le\frac{1}{4}\)

\(\Rightarrow B\ge1+\frac{2}{\frac{1}{4}}=9\)

Vậy B có GTNN bằng 9 khi x = y = \(\frac{1}{2}\)

Cho y ở đề bài làm gì trong khi biểu thức ở vế trái bên dưới ko có y?

à là \(\frac{8x}{y}\)đó

Ta có: \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+1+1+\frac{1}{x^2y^2}\)\(\Rightarrow\frac{x^4y^4+2x^2y^2+1}{x^2y^2}=\frac{\left(x^2y^2+1\right)^2}{x^2y^2}=\left(xy+\frac{1}{xy}\right)^2\)\(Tac\text{ó}:xy+\frac{1}{xy}=xy+\frac{1}{16xy}+\frac{15}{16xy}\)\(\text{ \text{áp} d\text{ụng} b\text{đ}t c\text{ô} si ta c\text{ó}: }\)

Áp dụng bddt cô si ta có :\(xy+\frac{1}{16xy}\ge2\sqrt{\frac{xy.1}{16xy}}=\frac{2.1}{4}=\frac{1}{2}\)

\(xy\le\frac{\left(x+y\right)^{2\Rightarrow}}{4}\Rightarrow xy\le\frac{1}{4}\Rightarrow\)\(\frac{1}{16xy}\ge\frac{4}{16}\Leftrightarrow\)\(\frac{15}{16xy}\le\frac{60}{16}=\frac{15}{4}\)\(\Rightarrow M=\left(xy+\frac{1}{xy}\right)^2\ge\left(\frac{1}{2}+\frac{15}{4}\right)^2=\left(\frac{17}{4}\right)^2=\frac{289}{16}\)

Dấu bằng xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)

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

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

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

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

\(=2+\left(x^2y^2+\frac{1}{256x^2y^2}\right)+\frac{255}{256x^2y^2}\)

Áp dụng BĐT Cauchy cho 2 số không âm:

\(x^2y^2+\frac{1}{256x^2y^2}\ge2\sqrt{\frac{x^2y^2}{256x^2y^2}}=\frac{1}{8}\)

C/m bđt phụ : \(1=\left(x+y\right)^2\ge4xy\)

\(\Rightarrow16x^2y^2\le1\Leftrightarrow256x^2y^2\le16\Leftrightarrow\frac{255}{256x^2y^2}\ge\frac{255}{16}\)

\(\Rightarrow A\ge2+\frac{1}{8}+\frac{255}{16}=\frac{289}{16}\)

(Dấu "="\(\Leftrightarrow\hept{\begin{cases}x^2y^2=\frac{1}{256x^2y^2}\\x-y=0\end{cases}}\Leftrightarrow x=y=\frac{1}{2}\))

thứ lỗi cho mk , mk không biết làm ; bài này khó quá

chuẩn k chỉnh

Áp dụng bđt AM-GM ta có

\(P\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2.\left(yz+1\right)^2.\left(zx+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}=A\)

  Ta có   \(A=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{zx+1}{z}\right)}=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)

Áp dụng bđt AM-GM ta có

\(A\ge3\sqrt[3]{8\sqrt{\frac{xyz}{xyz}}}=3.2=6\)

\(\Rightarrow P\ge6\)

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

Làm tiếp bài ღ๖ۣۜLinh's ๖ۣۜLinh'sღ] ★we are one★ chớ hình như bị ngược dấu ó.Do mình gà nên chỉ biết cô si mù mịt thôi ạ

\(3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)

\(=3\sqrt[3]{\left(y+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}\right)\left(z+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4y}\right)\left(x+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4z}\right)}\)

\(\ge3\sqrt[3]{5\sqrt[5]{\frac{y}{256x^4}}\cdot5\sqrt[5]{\frac{z}{256y^4}}\cdot5\sqrt[5]{\frac{x}{256z^4}}}\)

\(=3\sqrt[3]{125\sqrt[5]{\frac{xyz}{256^3\left(xyz\right)^4}}}\)

\(=15\sqrt[3]{\sqrt[5]{\frac{1}{256^3\left(xyz\right)^3}}}\)

\(\ge15\sqrt[15]{\frac{1}{256^3\cdot\left(\frac{x+y+z}{3}\right)^9}}\)

\(\ge15\sqrt[15]{\frac{1}{256^3\cdot\frac{1}{2^9}}}=\frac{15}{2}\)

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