Cho x,y là hai số thực thỏa mản x.y=1. Chứng minh \(\frac{4}{\left(x+y\right)^2}\) + \(^{x^2}\)+ \(y^2\)\(\ge3\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{4}{\left(x+y\right)^2}+x^2+y^2\)
\(=\frac{4}{\left(x+y\right)^2}+\left(x+y\right)^2-2xy\)
\(=\frac{4}{\left(x+y\right)^2}+\frac{\left(x+y\right)^2}{4}+\frac{3}{4}\left(x+y\right)^2-2\)
\(=\frac{4}{\left(x+y\right)^2}+\frac{\left(x+y\right)^2}{4}+\frac{3}{4}\left(x^2+y^2\right)+\frac{3}{4}2xy-2\)
\(=\frac{4}{\left(x+y\right)^2}+\frac{\left(x+y\right)^2}{4}+\frac{3}{4}\left(x^2+y^2\right)+\frac{3}{2}-2\)
Áp dụng bất đẳng thức Cauchy:
\(A\ge2\sqrt{\frac{4}{\left(x+y\right)^2}\frac{\left(x+y\right)^2}{4}}+\frac{3}{4}2\sqrt{x^2y^2}+\frac{3}{2}-2\)
\(A\ge2+\frac{3}{2}+\frac{3}{2}-2\)
\(A\ge3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{x^2}{y^2}+\frac{y^2}{x^2}+4\ge3\left(\frac{x}{y}+\frac{y}{x}\right)\)
\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}\right)^2-2+4-3\left(\frac{x}{y}+\frac{y}{x}\right)\ge0\)
\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}\right)^2-3\left(\frac{x}{y}+\frac{y}{x}\right)+2\ge0\)
\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}-1\right)\left(\frac{x}{y}+\frac{y}{x}+1\right)-3\left(\frac{x}{y}+\frac{y}{x}-1\right)\ge0\)
\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}-1\right)\left(\frac{x}{y}+\frac{y}{x}+2\right)\ge0\left(1\right)\)
Đến đây có 2 cách giải quyết
Cách 1:
\(\left(1\right)\Leftrightarrow\frac{x^2-xy+y^2}{xy}\cdot\frac{\left(x+y\right)^2}{xy}\ge0\)
\(\Leftrightarrow\frac{\left(x+y\right)^2\left(x^2-xy+y^2\right)}{x^2y^2}\ge0\)
\(\Leftrightarrow\frac{\left(x+y\right)^2\left[\left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4}\right]}{x^2y^2}\ge0\left(true!!!\right)\)
Cách 2 là đặt ẩn:)
Đặt \(\frac{x}{y}+\frac{y}{x}=t\Rightarrow t^2=\left(\frac{x}{y}+\frac{y}{x}\right)^2\ge4\cdot\frac{x}{y}\cdot\frac{y}{x}=4\)
\(\Rightarrow\left|t\right|\ge2\)
Khi đó ta có:
\(\left(t+1\right)\left(t-2\right)\ge0\)
Nếu \(t\ge2\Rightarrow t+1>0;t-2\ge0\Rightarrow\left(t+1\right)\left(t-2\right)\ge0\)
Nếu \(t\le-2\Rightarrow t+1< 0;t-2< 0\Rightarrow\left(t+1\right)\left(t-2\right)>0\)
=> đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-\left(y-1\right)}{\left(y-1\right)\left(y^2+y+1\right)}+\frac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\) \(+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=-\frac{1}{y^2+y+1}+\frac{1}{x^2+x+1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-\left(x^2+x+1\right)+y^2+y+1}{\left(y^2+y+1\right)\left(x^2+x+1\right)}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-\left(x^2-y^2\right)-\left(x-y\right)}{x^2y^2+x^2y+xy^2+x^2+y^2+xy+x+y+1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-\left(x-y\right)\left(x+y\right)-\left(x-y\right)}{x^2y^2+xy\left(x+y\right)+xy+x^2+y^2+2}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-\left(x-y\right)\left(x+y+1\right)}{x^2y^2+2xy+x^2+y^2+2}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-2\left(x-y\right)}{x^2y^2+\left(x+y\right)^2+2}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-2\left(x-y\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT=\left(\frac{1+x}{2}\right)^n+\left(\frac{1+y}{2}\right)^n+\left(\frac{1+z}{2}\right)^n\)
\(VT\ge\left(\frac{2\sqrt{x}}{2}\right)^n+\left(\frac{2\sqrt{y}}{2}\right)^n+\left(\frac{2\sqrt{z}}{2}\right)^n\)
\(VT\ge x^{\frac{n}{2}}+y^{\frac{n}{2}}+z^{\frac{n}{2}}\ge3\sqrt[3]{\left(xyz\right)^{\frac{n}{2}}}=3\)
Dấu "=" xảy ra khi \(x=y=z=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3+y^3+3xy\left(x+y\right)+\dfrac{1}{27}-3xy\left(x+y\right)-xy=0\)
\(\Leftrightarrow\left(x+y\right)^3+\dfrac{1}{27}-3xy\left(x+y+\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left(x+y+\dfrac{1}{3}\right)\left[\left(x+y\right)^2-\dfrac{1}{3}\left(x+y\right)+\dfrac{1}{9}\right]-3xy\left(x+y+\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow x^2+y^2-xy-\dfrac{1}{3}\left(x+y\right)+\dfrac{1}{9}=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-\dfrac{1}{3}\right)^2+\left(y-\dfrac{1}{3}\right)^2=0\)
\(\Leftrightarrow x=y=\dfrac{1}{3}\Rightarrow P=...\)
\(\frac{4}{\left(x+y\right)^2}+x^2+y^2=\frac{4}{x^2+y^2+2xy}+\left(x^2+y^2+1\right)-1\)
\(=\frac{4}{x^2+y^2+1}+\left(x^2+y^2+1\right)-1\)( vì xy=1)
\(\ge2\sqrt{\frac{4}{x^2+y^2+1}.\left(x^2+y^2+1\right)}-1\)(áp dụng bđt cô si )
\(=2\sqrt{4}-1=3\)
---> đpcm
cho mình hỏi là vì sao mà 2xy lại bằng 1 vậy?