Câu hỏi của Phạm Thu Hương

Question

Cho 2 số dương x,y thỏa mãn x+y=1 Tìm GTNN của biểu thức $\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)$

Phan Quỳnh · Accepted Answer

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}$$\Rightarrow M=\left(xy+\frac{1}{xy}\right)^2\ge\left(\frac{1}{2}+\frac{15}{4}\right)^2=\left(\frac{17}{4}\right)^2=\frac{289}{16}$Dấu bằng xảy ra $\Leftrightarrow x=y=\frac{1}{2}$Câu trả lời: Ta có: $M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+1+1+\frac{1}{x^2y^2}$$\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 $\Leftrightarrow x=y=\frac{1}{2}$

Kiệt Nguyễn · Answer

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}$)Câu trả lời: 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}$)

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