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10 tháng 5 2019
Chứng minh BĐT phụ:
\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)
Giờ thì chứng minh thôi:3
Áp dụng BĐT Cauchy-schwarz dạng engel ta có:
\(P=\left(2x+\frac{1}{x}\right)^2+\left(2y+\frac{1}{y}\right)^2\)
\(\ge\frac{\left(2x+\frac{1}{x}+2y+\frac{1}{y}\right)^2}{2}\)
\(\ge\frac{\left(2x+2y+\frac{4}{x+y}\right)^2}{2}\)
\(=\frac{\left[2\left(x+y\right)+\frac{4}{1}\right]^2}{2}\)
\(=8\)
Dấu "=" xảy ra khi và chỉ khi \(x=y=\frac{1}{2}\)
Vậy \(P_{min}=8\Leftrightarrow x=y=\frac{1}{2}\)
26 tháng 5 2019
Bài này bạn làm đúng rồi nhưng mà bạn bị nhầm phép tính: \(\frac{\left[2\left(x+y\right)+\frac{4}{1}\right]^2}{2}=18\)
=> Min P=18
25 tháng 3 2021
xin nhá xin nhá =))
Áp dụng bất đẳng thức Cauchy-Schwarz và giả thiết x+y=1 ta có :
\(P=\left(2x+\frac{1}{x}\right)^2+\left(2y+\frac{1}{y}\right)^2\ge\frac{\left(2x+\frac{1}{x}+2y+\frac{1}{y}\right)^2}{2}=\frac{\left[2\left(x+y\right)+\left(\frac{1}{x}+\frac{1}{y}\right)\right]^2}{2}\ge\frac{\left(2+\frac{4}{x+y}\right)^2}{2}=\frac{\left(2+4\right)^2}{2}=18\)
Đẳng thức xảy ra <=> x=y=1/2
Vậy ...
TA
22 tháng 8 2017
\(B=\left[\left(\frac{x}{y}-\frac{y}{x}\right):\left(x-y\right)-2.\left(\frac{1}{y}-\frac{1}{x}\right)\right]:\frac{x-y}{y}\)
\(=\left[\frac{x^2-y^2}{xy}.\frac{1}{x-y}-2.\frac{x-y}{xy}\right].\frac{y}{x-y}\)
\(=\left(\frac{\left(x-y\right)\left(x+y\right)}{xy.\left(x-y\right)}-\frac{2.\left(x-y\right)}{xy}\right).\frac{y}{x-y}\)
\(=\left(\frac{x+y}{xy}-\frac{2x-2y}{xy}\right).\frac{y}{x-y}=\frac{x+y-2x+2y}{xy}.\frac{y}{x-y}=\frac{y.\left(3y-x\right)}{xy.\left(x-y\right)}=\frac{3y-x}{x.\left(x-y\right)}\)
\(C=\left(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{2y^2}{y-x}\right):\frac{2y}{x-y}\)
\(=\left(\frac{x+y}{2.\left(x-y\right)}-\frac{x-y}{2.\left(x+y\right)}+\frac{2y^2}{x-y}\right).\frac{x-y}{2y}\)
\(=\frac{\left(x+y\right)^2-\left(x-y\right)^2+2.2y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{\left(x+y+x-y\right)\left(x+y-x+y\right)+4y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{4xy+4xy^2+4y^3}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}=\frac{4y.\left(x+xy+y^2\right).\left(x-y\right)}{4y.\left(x-y\right)\left(x+y\right)}=\frac{x+xy+y^2}{x+y}\)
\(D=3x:\left\{\frac{x^2-y^2}{x^3+y^3}.\left[\left(x-\frac{x^2+y^2}{y}\right):\left(\frac{1}{x}-\frac{1}{y}\right)\right]\right\}\)
\(=3x:\left\{\frac{\left(x+y\right)\left(x-y\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}.\left[\frac{xy-x^2-y^2}{y}:\frac{y-x}{xy}\right]\right\}\)
\(=3x:\left[\frac{x-y}{x^2-xy+y^2}.\left(\frac{xy-x^2-y^2}{y}.\frac{xy}{y-x}\right)\right]\)
\(=3x:\left(\frac{x-y}{x^2-xy+y^2}.\frac{xy.\left(x^2-xy+y^2\right)}{y.\left(x-y\right)}\right)\)
\(=3x:\frac{xy.\left(x-y\right)\left(x^2-xy+y^2\right)}{y.\left(x-y\right)\left(x^2-xy+y^2\right)}=3x:x=3\)
\(E=\frac{2}{x.\left(x+1\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+3\right)}\)
\(=2.\left(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)\)
\(=2.\frac{\left(x+2\right)\left(x+3\right)+x.\left(x+3\right)+x.\left(x+1\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{x^2+2x+3x+6+x^2+3x+x^2+x}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{3x^2+9x+6}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=2.\frac{3.\left(x^2+3x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x^2+x+2x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6.\left[x.\left(x+1\right)+2.\left(x+1\right)\right]}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x+1\right)\left(x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6}{x.\left(x+3\right)}\)
9 tháng 12 2018
\(A\)xác định \(\Leftrightarrow x^2y^2+1+\left(x^2-y\right)\left(1-y\right)\ne0\)
\(\Leftrightarrow x^2y^2+1+x^2-x^2y-y+y^2\ne0\)
\(\Leftrightarrow\left(x^2y^2+y^2\right)+\left(x^2+1\right)-\left(x^2y+y\right)\ne0\)
\(\Leftrightarrow y^2\left(x^2+1\right)+\left(x^2+1\right)-y\left(x^2+1\right)\ne0\)
\(\Leftrightarrow\left(x^2+1\right)\left(y^2-y+1\right)\ne0\)
\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\)
Ta có: \(\hept{\begin{cases}x^2+1>0\forall x\\\left(y-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall y\end{cases}}\)\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]>0\forall x;y\)
\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\forall x;y\)
\(\Leftrightarrow A\ne0\forall x;y\)
khai triển ra còn 4x^2+4y^2+1/x^2+1/y^2+8 =4(x^2+y^2)+(1/x^2+1/y^2)+8
>/ 4.(x+y)^2/2+8/(x+y)^2+8=18
"=" khi x=y=1/2
Đặt \(2x+\frac{1}{x}=a;2y+\frac{1}{y}=b\)
Ta có \(a^2+b^2>=2ab=>2\left(a^2+b^2\right)>=a^2+b^2+2ab=\left(a+b\right)^2\)
=>\(a^2+b^2>=\frac{\left(a+b\right)^2}{2}\)
Ta cần tìm giá trị nhỏ nhất của a+b
ta có \(a+b=2x+\frac{1}{x}+2y+\frac{1}{y}=2\left(x+y\right)+\frac{1}{x}+\frac{1}{y}=2+\frac{1}{x}+\frac{1}{y}\)
Áp dụng BĐT cauchy \(\frac{1}{x}+\frac{1}{y}>=\frac{4}{x+y}\)
=>\(a+b>=2+\frac{4}{x+y}=6\)
=>a\(a^2+b^2>=\frac{6^2}{2}=18\)
=>Min \(\left(2x+\frac{1}{x}\right)^2+\left(2y+\frac{1}{y}\right)^2\)=18
Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)