Cho x+y+z=1 Tính GTLN của P=2xy+xz+yz
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Dự đoán của chúa Pain x=y=z=1/3
áp dụng bất đẳng thức cô si ta có:
\(2xy\le2\left(\frac{x+y}{2}\right)^2\)
\(yz\le\left(\frac{y+z}{2}\right)^2\)
\(xz\le\left(\frac{z+x}{2}\right)^2\)
( vì X=Y=Z dự đoán của chúa pain) suy ra x+y=2x..ta được :
\(P\le2\left(\frac{x+y}{2}\right)^2+\left(\frac{y+z}{2}\right)^2+\left(\frac{z+x}{2}\right)^2\Leftrightarrow2x^2+y^2+z^2\)
\(P\le2x^2+y^2+z^2\Leftrightarrow P\le\frac{1}{3}\Leftrightarrow P\le\frac{2}{9}+\frac{1}{9}+\frac{1}{9}\Leftrightarrow P\le\frac{4}{9}\)
Vậy Max của P là 4/9 dâu = xảy ra khi x=y=z=1/3 đúng như dự đoán của chúa pain . chúa pain vô cmm nó địch :))
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\(2x+2y+z=4\Rightarrow z=4-2x-2y\)
Ta có: \(A=2xy+yz+xz\)
\(=2xy+y\left(4-2x-2y\right)+x\left(4-2x-2y\right)\)
\(=2xy+4y-2xy-2y^2+4x-2x^2-2xy\)
\(=4y-2xy-2y^2+4x-2x^2\)
\(\Rightarrow2A=8y-4xy-4y^2+8x-4x^2\)
\(=-4x^2-4x\left(y-2\right)-4y^2+8y\)
\(=-4x^2-2.x.2\left(y-2\right)-\left(y-2\right)^2+\left(y-2\right)^2-4y^2+8y\)
\(=-\left[4x^2+2.x.2\left(y-2\right)+\left(y-2\right)^2\right]+\left(y-2\right)^2-4y^2+8y\)
\(=-\left(2x+y-2\right)^2+y^2-4y+4-4x^2+8y\)
\(=-\left(2x+y-2\right)^2-3y^2+4y+4\)
\(=-\left(2x+y-2\right)^2-3\left(y^2-2.\frac{2}{3}y+\frac{4}{9}-\frac{4}{9}-\frac{4}{3}\right)\)
\(=-\left(2x+y-2\right)^2-3\left(y-\frac{2}{3}\right)^2+\frac{16}{3}\)
\(=\frac{16}{3}-\left[\left(2x+y-2\right)^2+3\left(y-\frac{2}{3}\right)^2\right]\)
Vì \(\left(2x+y-2\right)^2\ge0;\left(y-\frac{2}{3}\right)^2\ge0\) Nên \(\frac{16}{3}-\left[\left(2x+y-2\right)^2+3\left(y-\frac{2}{3}\right)^2\right]\le\frac{16}{3}\)
\(\Rightarrow A\le\frac{16}{3}:2=\frac{8}{3}\)
Dấu "=" xảy ra <=>\(\hept{\begin{cases}y-\frac{2}{3}=0\\2x+y-2=0\\z=4-2x-2y\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-y+2}{2}\\y=\frac{2}{3}\\z=4-2x-2y\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=\frac{2}{3}\\z=\frac{4}{3}\end{cases}}}\)
Vậy AMax = 8/3 khi và chỉ khi x = y = 2/3 và z = 4/3
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\(A=2xy+yz+xz\)
\(=2xy+y\left(4-2x-2y\right)+x\left(4-2x-2y\right)\)
\(=-2x^2-2xy+4x-2y^2+4y\)
\(=\left[-\left(x^2+2xy+y^2\right)+\dfrac{8}{3}\left(x+y\right)-\dfrac{16}{9}\right]-\left(x^2-\dfrac{4}{3}x+\dfrac{4}{9}\right)-\left(y-\dfrac{4}{3}y+\dfrac{4}{9}\right)+\dfrac{8}{3}\)\(=-\left(x+y-\dfrac{4}{3}\right)^2-\left(x-\dfrac{2}{3}\right)^2-\left(y-\dfrac{2}{3}\right)^2+\dfrac{8}{3}\le\dfrac{8}{3}\)
Vậy \(A_{max}=\dfrac{8}{3}\) tại \(\left\{{}\begin{matrix}x=y=\dfrac{2}{3}\\z=\dfrac{4}{3}\end{matrix}\right.\)
z = 4-2(x+y)
=> A= 2xy + y[4-2(x+y)] + x[4-2(x+y)]
=\(2xy+4y-2xy-2y^2+4x-2x^2-2xy\)
= \(-\left(y^2-4y+4\right)-\left(x^2-4x+4\right)-\left(y^2+2xy+x^2\right)+8\)
=\(8-\left[\left(y-2\right)^2+\left(x-2\right)^2-\left(y-x\right)^2\right]\le8\forall x,y\)
vậy GTLN của A là 8 khi x=y=2
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Ta có \(\frac{x+2xy+1}{x+xy+xz+1}=\frac{x+2xy+xyz}{x+xy+xz+xyz}=\frac{1+2y+yz}{\left(y+1\right)\left(z+1\right)}\)
Tương tự => \(M=\frac{1+2y+yz}{\left(y+1\right)\left(z+1\right)}+\frac{1+2z+zx}{\left(1+x\right)\left(z+1\right)}+\frac{1+2x+xy}{\left(1+x\right)\left(y+1\right)}\)
=> \(M=\frac{\left(1+2y+yz\right)\left(1+x\right)+\left(1+2z+zx\right)\left(1+y\right)+\left(1+2x+xy\right)\left(1+z\right)}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)
=>\(M=\frac{6+3\left(x+y+z\right)+3\left(xy+yz+xz\right)}{2+\left(x+y+z\right)+\left(xy+yz+xz\right)}=3\)
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\(\left(x^3+3x^2y+3xy^2+y^3-z^3\right):\left(x+y-z\right)\\ =\left[\left(x+y\right)^3-z^3\right]:\left(x+y-z\right)\\ =\left(x+y-z\right)\left[\left(x+y\right)^2+z\left(x+y\right)+z^2\right]:\left(x+y-z\right)\\ =x^2+2xy+y^2+xz+yz+z^2\)
Vậy chọn A
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