Câu hỏi của Nguyễn Thị Mát

Question

Cho 3 số thực dương x,y,z có tích bằng 1.Chứng minh rằng:

$\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{y^3}{\left(1+z\right)\left(1+x\right)}+\frac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\frac{3}{4}$

Kudo Shinichi · Accepted Answer

Ta có : $\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{1+y}{8}+\frac{1+z}{8}\ge\frac{3x}{4}$$\Rightarrow\frac{x^3}{\left(1+y\right)\left(1+z\right)}\ge\frac{6x-y-z-2}{8}\left(1\right)$Tương tự ta có : $\hept{\begin{cases}\frac{y^3}{\left(1+z\right)\left(1+x\right)}\ge\frac{6y-z-x-2}{8}\left(2\right)\\frac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\frac{6z-x-y-2}{8}\left(3\right)\end{cases}}$Từ (1) , (2) , (3)$\Rightarrow\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{y^3}{\left(1+z\right)\left(1+x\right)}+\frac{z^3}{\left(1+x\right)\left(1+y\right)}$$\ge\frac{6x-y-z-2}{8}+\frac{6y-z-x-2}{8}+\frac{6z-x-y-2}{8}$$=\frac{1}{2}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{2}-\frac{3}{4}=\frac{3}{4}$Chúc bạn học tốt !!!Câu trả lời: Ta có : $\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{1+y}{8}+\frac{1+z}{8}\ge\frac{3x}{4}$$\Rightarrow\frac{x^3}{\left(1+y\right)\left(1+z\right)}\ge\frac{6x-y-z-2}{8}\left(1\right)$Tương tự ta có : $\hept{\begin{cases}\frac{y^3}{\left(1+z\right)\left(1+x\right)}\ge\frac{6y-z-x-2}{8}\left(2\right)\\frac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\frac{6z-x-y-2}{8}\left(3\right)\end{cases}}$Từ (1) , (2) , (3)$\Rightarrow\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{y^3}{\left(1+z\right)\left(1+x\right)}+\frac{z^3}{\left(1+x\right)\left(1+y\right)}$$\ge\frac{6x-y-z-2}{8}+\frac{6y-z-x-2}{8}+\frac{6z-x-y-2}{8}$$=\frac{1}{2}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{2}-\frac{3}{4}=\frac{3}{4}$Chúc bạn học tốt !!!

Lê Tài Bảo Châu · Answer

Áp dụng bđt AM-GM ta có: $\hept{\begin{cases}\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{1+y}{8}+\frac{1+z}{8}\ge3\sqrt[3]{\frac{x^3}{\left(1+y\right)\left(1+z\right)}.\frac{1+y}{8}.\frac{1+z}{8}}=\frac{3x}{4}\left(1\right)\\frac{y^3}{\left(1+z\right)\left(1+x\right)}+\frac{1+z}{8}+\frac{1+x}{8}\ge3\sqrt[3]{\frac{y^3}{\left(1+z\right)\left(1+x\right)}.\frac{1+z}{8}.\frac{1+x}{8}}=\frac{3y}{4}\left(2\right)\\frac{z^3}{\left(1+x\right)\left(1+y\right)}+\frac{1+x}{8}+\frac{1+y}{8}\ge3\sqrt[3]{\frac{z^3}{\left(1+x\right)\left(1+y\right)}.\frac{1+x}{8}.\frac{1+y}{8}}=\frac{3z}{4}\left(3\right)\end{cases}}$Lấy $\left(1\right)+\left(2\right)+\left(3\right)$ta được: $P+\frac{3+x+y+z}{4}\ge\frac{3\left(x+y+z\right)}{4}$$\Leftrightarrow P\ge\frac{3\left(x+y+z\right)}{4}-\frac{3+x+y+z}{4}$$\Leftrightarrow P\ge\frac{2\left(x+y+z\right)-3}{4}\left(1\right)$Áp dụng bdt AM-GM ta có:$x+y+z\ge3\sqrt[3]{xyz}=3$Thay vào (1) ta được:$P\ge\frac{2.3-3}{4}$$\Rightarrow P\ge\frac{3}{4}$Dấu"="xảy ra $\Leftrightarrow x=y=z$Câu trả lời: Áp dụng bđt AM-GM ta có: $\hept{\begin{cases}\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{1+y}{8}+\frac{1+z}{8}\ge3\sqrt[3]{\frac{x^3}{\left(1+y\right)\left(1+z\right)}.\frac{1+y}{8}.\frac{1+z}{8}}=\frac{3x}{4}\left(1\right)\\frac{y^3}{\left(1+z\right)\left(1+x\right)}+\frac{1+z}{8}+\frac{1+x}{8}\ge3\sqrt[3]{\frac{y^3}{\left(1+z\right)\left(1+x\right)}.\frac{1+z}{8}.\frac{1+x}{8}}=\frac{3y}{4}\left(2\right)\\frac{z^3}{\left(1+x\right)\left(1+y\right)}+\frac{1+x}{8}+\frac{1+y}{8}\ge3\sqrt[3]{\frac{z^3}{\left(1+x\right)\left(1+y\right)}.\frac{1+x}{8}.\frac{1+y}{8}}=\frac{3z}{4}\left(3\right)\end{cases}}$Lấy $\left(1\right)+\left(2\right)+\left(3\right)$ta được: $P+\frac{3+x+y+z}{4}\ge\frac{3\left(x+y+z\right)}{4}$$\Leftrightarrow P\ge\frac{3\left(x+y+z\right)}{4}-\frac{3+x+y+z}{4}$$\Leftrightarrow P\ge\frac{2\left(x+y+z\right)-3}{4}\left(1\right)$Áp dụng bdt AM-GM ta có:$x+y+z\ge3\sqrt[3]{xyz}=3$Thay vào (1) ta được:$P\ge\frac{2.3-3}{4}$$\Rightarrow P\ge\frac{3}{4}$Dấu"="xảy ra $\Leftrightarrow x=y=z$

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