cho\(\hept{\begin{cases}x,y>0\\x^3+y^4\le x^2y^3\end{cases}}\)chứng minh rằng
a) \(x^2+y^3\le x^2+y^2\)
b) \(x^2+y^3\le x+y^2\)
Cho x, y, z thỏa mãn:
\(\hept{\begin{cases}0\le x,y,z\le1\\2x+y\le2\end{cases}}\)
Chứng minh \(2x^2+y^2\le\frac{3}{2}\)
Cho x, y là các số thỏa mãn đồng thời : \(\hept{\begin{cases}0\le x\le y\le1\\2x+y\le2\end{cases}.}\)
Chứng minh bất đẳng thức : \(2x^2+y^2\le\frac{3}{2}.\)
Từ \(0\le x\le y\le1\) và \(2x+y\le2\Rightarrow2x^2+xy\le2x\)(nhân cả 2 vế với \(x\ge0\))
\(\left(y-x\right)y\le y-x\)(nhân cả 2 vế của \(0\le y\le1\)với \(y-x\ge0\)(do \(x\le y\))
Cộng từng vế ta có :
\(2x^2+xy+\left(y-x\right)y\le2x+y-x\)
\(\Leftrightarrow2x^2+y^2\le x+y\)
\(\Leftrightarrow\left(2x^2+y^2\right)^2\le\left(x+y\right)^2\)
Mặt khác \(\left(x+y\right)^2=\left(\frac{1}{\sqrt{2}}.\sqrt{2}x+1.y\right)^2\le\left(\frac{1}{2}+1\right)\left(2x^2+y^2\right)\)(bất đẳng thức Bunhiacopxki)
\(\Rightarrow\left(2x^2+y^2\right)^2\le\frac{3}{2}\left(2x^2+y^2\right).\)
\(\Leftrightarrow2x^2+y^2\le\frac{3}{2}.\)(đpcm)
Chúc học tốt
Cho x, y t/m \(\hept{\begin{cases}\text{x, y }\varepsilon R\\0\le x;y\le\frac{1}{2}\end{cases}}\). CMR: \(\frac{\sqrt{x}}{1+y}+\frac{\sqrt{y}}{1+x}\le\frac{2\sqrt{2}}{3}\)
Cho \(\hept{\begin{cases}x,y\in R\\0\le x,y\le\frac{1}{2}\end{cases}}\)
CMR : \(\frac{\sqrt{x}}{1+y}+\frac{\sqrt{y}}{1+x}\le\frac{2\sqrt{2}}{3}\)
Từ gt => \(\hept{\begin{cases}\left(\frac{1}{\sqrt{2}}-\sqrt{x}\right)\left(\frac{1}{\sqrt{2}}-\sqrt{y}\right)\ge0\Leftrightarrow\sqrt{x}+\sqrt{y}\le\frac{\sqrt{2}}{2}+\sqrt{2}\sqrt{xy}\left(1\right)\\x\sqrt{x}\le x\cdot\frac{1}{\sqrt{2}};y\sqrt{y}\le y\cdot\frac{1}{\sqrt{2}}\Rightarrow x\sqrt{x}+y\sqrt{y}\le\frac{1}{\sqrt{2}}\left(x+y\right)\left(2\right)\end{cases}}\)
Lại có \(\hept{\begin{cases}\sqrt{xy}\le xy+\frac{1}{4}\\\sqrt{xy}\le\frac{x+y}{2}\end{cases}\Rightarrow\hept{\begin{cases}\frac{2\sqrt{2}}{3}\sqrt{xy}\le\frac{2\sqrt{2}}{3}\left(xy+\frac{1}{4}\right)\left(3\right)\\\frac{\sqrt{2}}{3}\sqrt{xy}\le\frac{\sqrt{2}}{6}\left(x+y\right)\left(4\right)\end{cases}}}\)
Từ (1)(2)(3)(4) ta có:\(x\sqrt{x}+y\sqrt{y}+\sqrt{x}+\sqrt{y}\le\frac{\sqrt{2}}{2}\left(x+y\right)+\frac{\sqrt{2}}{2}+\frac{2\sqrt{2}}{3}\left(xy+\frac{1}{4}\right)+\frac{\sqrt{2}}{6}\left(x+y\right)\)
\(\le\frac{2\sqrt{2}}{3}\left(1+x+y+xy\right)\)
=> \(VT=\frac{\sqrt{x}}{1+y}+\frac{\sqrt{y}}{1+x}=\frac{x\sqrt{x}+y\sqrt{y}+\sqrt{x}+\sqrt{y}}{1+x+y+xy}\le\frac{2\sqrt{2}}{3}\)
Dấu "=" xảy ra <=> x=y=\(\frac{1}{2}\)
Cho x, y t/m \(\hept{\begin{cases}\text{x, y }\varepsilon R\\0\le x;y\le\frac{1}{2}\end{cases}}\). CMR: \(\frac{\sqrt{x}}{1+y}+\frac{\sqrt{y}}{1+x}\le\frac{2\sqrt{2}}{3}\)
Từ gt => \(\hept{\begin{cases}\left(\frac{1}{\sqrt{2}}-x\right)\left(\frac{1}{\sqrt{2}}-y\right)\ge0\Leftrightarrow\sqrt{x}+\sqrt{y}\le\frac{\sqrt{2}}{2}+\sqrt{2}\sqrt{xy}\left(1\right)\\x\sqrt{x}\le x\cdot\frac{1}{\sqrt{2}};y\sqrt{y}\le y\cdot\frac{1}{\sqrt{2}}\Rightarrow x\sqrt{x}+y\sqrt{y}\le\frac{1}{\sqrt{2}}\left(x+y\right)\left(2\right)\end{cases}}\)
Lại có \(\hept{\begin{cases}\sqrt{xy}\le xy+\frac{1}{4}\\\sqrt{xy}\le\frac{x+y}{2}\end{cases}\Rightarrow\hept{\begin{cases}\frac{2\sqrt{2}}{3}\sqrt{xy}\le\frac{2\sqrt{2}}{3}\left(xy+\frac{1}{4}\right)\left(3\right)\\\frac{\sqrt{2}}{3}\sqrt{xy}\le\frac{\sqrt{2}}{6}\left(x+y\right)\left(4\right)\end{cases}}}\)
Từ (1)(2)(3) và (4) ta có:
\(x\sqrt{x}+y\sqrt{y}+\sqrt{x}+\sqrt{y}\le\frac{\sqrt{2}}{2}\left(x+y\right)+\frac{\sqrt{2}}{2}+\frac{2\sqrt{2}}{3}\left(xy+\frac{1}{4}\right)+\frac{\sqrt{2}}{6}\left(x+y\right)\)
\(\le\frac{2\sqrt{2}}{3}\left(1+x+y+xy\right)\)
=> \(VT=\frac{\sqrt{x}}{1+y}+\frac{\sqrt{y}}{1+x}=\frac{x\sqrt{x}+y\sqrt{y}+\sqrt{x}+\sqrt{y}}{1+x+y+xy}\le\frac{2\sqrt{2}}{3}\)
Dấu "=" xảy ra <=> \(x=y=\frac{1}{2}\)
Biểu diễn hình học tập nghiệm của các bất phương trình bậc nhất hai ẩn sau:
a,\(\hept{\begin{cases}2x-1\le0\\-3x+5\le0\end{cases}}\)
b,\(\hept{\begin{cases}3-y< 0\\2x-3y+1>0\end{cases}}\)
c,\(\hept{\begin{cases}x-2y< 0\\x+3y>-2\end{cases}}\)
d,\(\hept{\begin{cases}3x-2y-6\ge0\\2\left(x-1\right)+\frac{3y}{2}\le4\\x\ge0\end{cases}}\)
e,\(\hept{\begin{cases}x-y>0\\x-3y\le-3\\x+y>5\end{cases}}\)
f,\(\hept{\begin{cases}x-3y< 0\\x+2y>-3\\y+x< 2\end{cases}}\)
Cho x,y,z thỏa mãn: \(\hept{\begin{cases}xy+yz+zx=1\\x^2+y^2+z^2=2\end{cases}}\). Cmr : \(\frac{-4}{3}\le x,y,z\le\frac{4}{3}\)
Ta có \(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)=4\)
=> \(\orbr{\begin{cases}x+y+z=2\\x+y+z=-2\end{cases}}\)
+ \(x+y+z=2\)
Thay vào Pt (1)
=> \(xy+z\left(2-z\right)=1\)
=> \(xy=\left(z-1\right)^2\)=> \(x,y,z\ge0\)( do \(x+y+z=2>0\))
Mà \(xy\le\left(\frac{x+y}{2}\right)^2=\left(\frac{2-z}{2}\right)^2\)
=> \(z-1\le\frac{2-z}{2}\)=> \(z\le\frac{4}{3}\)
Hoàn toàn TT => \(x,y,z\le\frac{4}{3}\)
+ \(x+y+z=-2\)
=> \(xy+z\left(-2-z\right)=1\)
=> \(xy=\left(z+1\right)^2\)=> \(x,y,z\le0\)( do \(x+y+z=-2< 0\))
Mà \(xy\le\left(\frac{x+y}{2}\right)^2=\left(\frac{-2-z}{2}\right)^2\)
=> \(\left(z+1\right)^2\le\left(\frac{z+2}{2}\right)^2\)
=> \(z+1\ge\frac{-z-2}{2}\)=> \(z\ge-\frac{4}{3}\)
TT => \(x,y,z\ge-\frac{4}{3}\)
Vậy \(-\frac{4}{3}\le x,y,z\le\frac{4}{3}\)
\(\hept{\begin{cases}x+y+z=5\\x^2+y^2+z^2=9\end{cases}}\)CMR:\(1\le x,y,z\le\frac{7}{3}\)
\(1,\hept{\begin{cases}\sqrt{x}+\sqrt{y}=3\\\sqrt{x+5}+\sqrt{y+3}=5\end{cases}}\)
\(2,\hept{\begin{cases}x\left(x+y+1\right)-3=0\\\left(x+y\right)^2-\frac{5}{x^2}+1=0\end{cases}}\)
\(3,\hept{\begin{cases}xy+x+y=x^2+2y^2\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\end{cases}}\)
\(4,\hept{\begin{cases}xy+x+1=7y\\x^2y^2+xy+1=13y^2\end{cases}}\)
\(5,\hept{\begin{cases}2y\left(x^2-y^2\right)=3x\\x\left(x^2+y^2\right)=10y\end{cases}}\)