cho\(x,y>0\) \(x+y\le1\)tìm min \(\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=\(\frac{x}{y}+\frac{y}{x}\)
Đặt \(\frac{x}{y}=a\left(a>0\right)\)
vì x,y>0 áp dụng bđt cô si
\(x+\frac{1}{y}\ge2\sqrt{\frac{x}{y}}\)
\(1\ge x+\frac{1}{y}\ge2\sqrt{\frac{x}{y}}\)
\(\frac{1}{4}\ge\frac{x}{y}\)
\(0< a\le\frac{1}{4}\)
Có A=\(a+\frac{1}{a}\left(với0< a\le\frac{1}{4}\right)\)
A=\(16a+\frac{1}{a}-15a\)
a>0 cô si
A\(\ge2\sqrt{16a\cdot\frac{1}{a}}-15\cdot\frac{1}{4}=\frac{17}{4}\)
D=XR x=y=1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x+y+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)=x+\frac{1}{4x}+y+\frac{1}{4y}+\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Áp dụng BĐT AM-GM: \(A\ge2\sqrt{\frac{x}{4x}}+2\sqrt{\frac{y}{4y}}+\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)=2+\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Áp dụng BĐT Schwarz dạng Engel: \(A\ge2+\frac{1}{4}.\frac{4}{x+y}\ge3\) (Do \(x+y\le1\))
Vậy Min A = 3. Dấu "=" xảy ra <=> x=y=1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
ap dung bdt cauchy schwarz ta co
\(\frac{\left(x-1\right)^2}{z}+\frac{\left(y-1\right)^2}{x}+\frac{\left(z-1\right)^2}{y}>=\frac{\left(x-1+z-1+y-1\right)^2}{x+y+z}=\frac{1}{2}\)
vay min=1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì a,b>0
A\(\ge2\sqrt{\frac{1}{x}\cdot\frac{1}{y}}\cdot\sqrt{1+x^2y^2}\)
A\(\ge2\sqrt{\frac{1+x^2y^2}{xy}}\)
A\(\ge2\sqrt{\frac{1}{xy}+xy}\)
Đặt xy=a, a>0
Ta cs xy\(\le\frac{\left(x+y\right)^2}{4}\le\frac{1^2}{4}=\frac{1}{4}\)
ĐK 0<a<\(\frac{1}{4}\)
\(\Leftrightarrow A\ge2\sqrt{\frac{1}{a}+a}\)
A\(\ge2\sqrt{16a+\frac{1}{a}-15a}\)
a>0, áp dụng bđt cô si
\(A\ge2\sqrt{2\sqrt{16a\cdot\frac{1}{a}}-\frac{15}{4}}\)
A\(\ge\sqrt{17}\)
Dấu = x ra a=b=0.5
![](https://rs.olm.vn/images/avt/0.png?1311)
2, rút gọn B=x^2/(y-1)+y^2/(x-1)
AM-GM : x^2/(y-1)+4(y-1) >/ 4x ; y^2/(x-1)+4(x-1) >/ 4y
=> B >/ 4x-4(y-1)+4y-4(x-1)=4x-4y+4+4y-4x+4=8
minB=8
Câu 1:
Áp dụng BĐT AM-GM ta có: \(x+1\ge2\sqrt{x}\)
\(\Rightarrow x+1+x+1\ge x+2\sqrt{x}+1\)
\(\Rightarrow2x+2\ge\left(\sqrt{x}+1\right)^2\left(1\right)\)
Tương tự cũng có: \(2y+2\ge\left(\sqrt{y}+1\right)^2\left(2\right)\)
Nhân theo vế của \(\left(1\right);\left(2\right)\) ta có:
\(\left(2x+2\right)\left(2y+2\right)\ge\left(\sqrt{x}+1\right)^2\left(\sqrt{y}+1\right)^2\ge16\)
\(\Rightarrow4\left(x+1\right)\left(y+1\right)\ge16\Rightarrow\left(x+1\right)\left(y+1\right)\ge4\)
Lại áp dụng BĐT AM-GM ta có:
\(\left(x+1\right)+\left(y+1\right)\ge2\sqrt{\left(x+1\right)\left(y+1\right)}\ge4\)
\(\Rightarrow x+y\ge2\). Giờ thì áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(A=\frac{x^2}{y}+\frac{y^2}{x}\ge\frac{\left(x+y\right)^2}{x+y}=x+y\ge2\)
Đẳng thức xảy ra khi \(x=y=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P\ge\frac{2}{\sqrt{xy}}\sqrt{1+x^2y^2}=2\sqrt{\frac{1+x^2y^2}{xy}}=2\sqrt{\frac{1}{xy}+xy}\)\(=2\sqrt{\frac{1}{16xy}+xy+\frac{15}{16xy}}\ge2\sqrt{\frac{1}{2}+\frac{15}{4\left(x+y\right)^2}}=\sqrt{17}.\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sử dụng BĐT Am-Gm ta có:
\(A=2\left(\frac{1}{x}+\frac{1}{y}\right)+\left(x+y\right)^2\ge4xy+\frac{4}{\sqrt{xy}}\)
\(\Rightarrow A\ge4xy+\frac{2}{\sqrt{xy}}+\frac{2}{\sqrt{xy}}\ge3\sqrt[3]{4xy.\frac{2}{\sqrt{xy}}.\frac{2}{\sqrt{xy}}}=6\sqrt[3]{2}\)
Dấu = xảy ra khi \(\hept{\begin{cases}x=y\\4xy=\frac{2}{\sqrt{xy}}\end{cases}}\Rightarrow x=y=\frac{1}{\sqrt[3]{2}}\)
\(A=\left(1+\frac{1}{x}\right)^2+\left(1+\frac{1}{y}\right)^2\)
Ta co:\(x+\frac{1}{x}=\left(\frac{1}{x}+4x\right)-3x\ge2\sqrt{\frac{1}{x}\cdot4x}-3x=4-3x\left(AM-GM\right)\)
Tuong tu:\(y+\frac{1}{y}=4-3y\)
Ta co:\(A\ge\left(4-3x\right)^2+\left(4-3y\right)^2\)
\(=16-24x+9x^2+16-24y+9y^2\)
\(=32-24\left(x+y\right)+9\left(x^2+y^2\right)\)
Ap dung bat dang thuc phu:\(\frac{\left(x+y\right)^2}{4}\le\frac{x^2+y^2}{2}\Rightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)
Khi do,ta co:
\(A\ge32-24\cdot1+9\cdot\frac{1}{2}=\frac{25}{2}\)
Dau bang xay ra khi va chi khi:\(x=y=\frac{1}{2}\)
P/S:E ko chac dau ah,e ms lm quen vs no thoi
\(VT\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\ge\frac{\left(4\left(x+y\right)+\frac{4}{x+y}-3\left(x+y\right)\right)^2}{2}\)
\(\ge\frac{\left(2.4-3.1\right)^2}{2}=\frac{25}{2}\)
đẳng thức xảy ra khi x = y = 1/2