Chứng minh rằng:
a) x2 - 2xy + 6y2 - 12x + 2y + 41 (lớn hơn hoặc bằng) 0
b) \(\dfrac{x^2}{y^2}\) + \(\dfrac{y^2}{x^2}\) - \(\dfrac{2x}{y}\) - \(\dfrac{2y}{x}\) + 3 > 0
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12 tháng 2 2022
\(\dfrac{2xy}{x^2+4y^2}+\dfrac{y^2}{3x^2+2y^2}\le\dfrac{3}{5}\)
<=> \(\left(\dfrac{2}{5}-\dfrac{2xy}{x^2+4y^2}\right)+\left(\dfrac{1}{5}-\dfrac{y^2}{3x^2+2y^2}\right)\ge0\)
<=> \(\dfrac{2x^2+8y^2-10xy}{x^2+4y^2}+\dfrac{3x^2+2y^2-5y^2}{3x^2+2y^2}\ge0\)
<=> \(\dfrac{2\left(x-4y\right)\left(x-y\right)}{x^2+4y^2}+\dfrac{3\left(x+y\right)\left(x-y\right)}{3x^2+2y^2}\ge0\)
<=> \(\left(x-y\right)\left[\dfrac{2\left(x-4y\right)}{x^2+4y^2}+\dfrac{3\left(x+y\right)}{3x^2+2y^2}\right]\ge0\) (1)
Xét \(\dfrac{2\left(x-4y\right)}{x^2+4y^2}+\dfrac{3\left(x+y\right)}{3x^2+2y^2}=\dfrac{2\left(x-4y\right)\left(3x^2+2y^2\right)+3\left(x+y\right)\left(x^2+4y^2\right)}{\left(x^2+4y^2\right)\left(3x^2+2y^2\right)}\)
= \(\dfrac{9x^3+16xy^2-21x^2y-4y^3}{\left(x^2+4y^2\right)\left(3x^2+2y^2\right)}=\dfrac{\left(x-y\right)\left(3x-2y\right)^2}{\left(x^2+4y^2\right)\left(3x^2+2y^2\right)}\)
(1) <=> \(\dfrac{\left(x-y\right)^2\left(3x-2y\right)^2}{\left(x^2+4y^2\right)\left(3x^2+2y^2\right)}\ge0\) (luôn đúng)
=> \(A\le\dfrac{3}{5}\)
Dấu "=" xảy ra \(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=\dfrac{2}{3}y\end{matrix}\right.\)
PD
1
TT
6 tháng 11 2017
\(\dfrac{2x^2+3xy+y^2}{2x^3+x^2y-2xy^2-y^3}=\dfrac{1}{x-y}\)
\(VT=\dfrac{2x^2+3xy+y^2}{2x^3+x^2y-2xy^2-y^3}\)
\(=\dfrac{2x^2+2xy+xy+y^2}{\left(2x^3+x^2y\right)+\left(-2xy^2-y^3\right)}\)
\(=\dfrac{\left(2x^2+2xy\right)+\left(xy+y^2\right)}{x^2\left(2x+y\right)-y^2\left(2x+y\right)}\)
\(=\dfrac{2x\left(x+y\right)+y\left(x+y\right)}{\left(x^2-y^2\right)\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)\left(x+y\right)}{\left(x^2-y^2\right)\left(2x+y\right)}\)
\(=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{1}{x-y}=VP\left(đpcm\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 3 2021
\(VT=\dfrac{x^2}{x^2+2xy+3zx}+\dfrac{y^2}{y^2+2yz+3xy}+\dfrac{z^2}{z^2+2zx+3yz}\)
\(VT\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+5xy+5yz+5zx}=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+3\left(xy+yz+zx\right)}\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\left(x+y+z\right)^2}=\dfrac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 4 2021
\(VT\le\dfrac{x}{2x+2y+2}+\dfrac{y}{2yz+2z+2}+\dfrac{z}{2z+2x+2}\)
Nên ta chỉ cần chứng minh: \(\dfrac{x}{x+y+1}+\dfrac{y}{y+z+1}+\dfrac{z}{z+x+1}\le1\)
\(\Leftrightarrow\dfrac{y+1}{x+y+1}+\dfrac{z+1}{y+z+1}+\dfrac{x+1}{z+x+1}\ge2\)
Thật vậy, ta có:
\(VT=\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(z+x+1\right)}+\dfrac{\left(y+1\right)^2}{\left(y+1\right)\left(x+y+1\right)}+\dfrac{\left(z+1\right)^2}{\left(z+1\right)\left(y+z+1\right)}\)
\(VT\ge\dfrac{\left(x+y+z+3\right)^2}{\left(x^2+y^2+z^2\right)+3\left(x+y+z\right)+xy+yz+zx+3}\)
\(VT\ge\dfrac{6\left(x+y+z\right)+2\left(xy+yz+zx\right)+12}{3\left(x+y+z\right)+xy+yz+zx+6}=2\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
\(a.x^2-2xy+6y^2-12x+2y+41\)
\(=x^2-2xy+y^2-12x+12y+36+5y^2-10y+5\)
\(=\left(x-y\right)^2-2.6\left(x-y\right)+36+5\left(y-1\right)^2\)
\(=\left(x-y-6\right)^2+5\left(y-1\right)^2\) ≥ \(0\)
\(b.\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}-\dfrac{2x}{y}-\dfrac{2y}{x}+3\)
\(=\dfrac{x^2}{y^2}-2.\dfrac{x}{y}+1+\dfrac{y^2}{x^2}-2.\dfrac{y}{x}+1+1\)
\(=\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2+1>0\)
Dấu "=" xảy ra khi............. LL hết lười =))