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áp dụng BĐT bunhiacopxki ta có
\(\left(1+2^2\right)\left(x^2+4y^2\right)\ge\left(x+4y\right)^2\)
<=> \(5\left(x^2+4y^2\right)\ge1\)
<=> \(x^2+4y^2\ge\dfrac{1}{5}\) (đpcm)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(1^2+2^2\right)\left(x^2+4y^2\right)\ge\left(x+4y\right)^2\)
\(\Rightarrow5\left(x^2+4y^2\right)\ge\left(x+4y\right)^2\)
\(\Rightarrow5\left(x^2+4y^2\right)\ge1^2=1\)
\(\Rightarrow5\left(x^2+4y^2\right)\ge\dfrac{1}{5}\)
Đẳng thức xảy ra khi \(x=y=\dfrac{1}{5}\)
1) a) Đặt biểu thức là A
\(A=2x^2+4y^2-4xy-4x-4y+2017\)
\(A=\left(x-2y\right)^2+x^2-4x-4y+2017\)
\(A=\left(x-2y\right)^2+2\left(x-2y\right)+x^2-6x+2017\)
\(A=\left(x-2y-1\right)^2+\left(x+3\right)^2+2008\)
Vậy: MinA=2008 khi x=-3; y=-2
3) a) \(A=\dfrac{1}{x^2+x+1}\)
\(B=x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(\Rightarrow B\ge\dfrac{3}{4}\Rightarrow A\ge\dfrac{4}{3}\)
Vậy MinA là \(\dfrac{4}{3}\) khi x=-0,5
\(A=\left(x^2+2x+1\right)+\left(y^2-6y+9\right)=\left(x+1\right)^2+\left(y-3\right)^2\)
Mà (x+1)^2>=0
(y-3)^2>=0
=> (x+1)^2+(y-3)^2>=0
Ta có: \(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}\ge\dfrac{2}{1+xy}\)
\(\Leftrightarrow\left(\dfrac{1}{1+x^2}-\dfrac{1}{1+y^2}\right)+\left(\dfrac{1}{1+y^2}-\dfrac{1}{xy}\right)\ge0\)
\(\Leftrightarrow\dfrac{xy-x^2}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{xy-y^2}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow\dfrac{x\left(y-x\right)}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{y\left(x-y\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(y-x\right)^2\left(xy-1\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)
BĐT cuối đúng vì x.y > 0 => đpcm
-(a - 3)2 = -(a2 - 6a + 9) = -a2 + 6a - 9
(x - 2)(x + 2) = x2 - 4
-(5 + 4y)(5 - 4y) = -(25 - 16y2) = -25 + 16y2
(\(\dfrac{1}{2}\)x + 2y)(\(\dfrac{1}{2}\)x - 2y) = \(\dfrac{1}{4}\)x2 - 4y2
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\) MTC: \(xy\left(x-2y\right)\left(x+2y\right)\)
\(=\dfrac{2x.y\left(x-2y\right)}{xy\left(x+2y\right)\left(x-2y\right)}+\dfrac{y.x\left(x+2y\right)}{xy\left(x-2y\right)\left(x+2y\right)}+\dfrac{4.xy}{xy\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-4xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y-2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) \(\dfrac{1}{x-y}+\dfrac{3xy}{y^3-x^3}+\dfrac{x-y}{x^2+xy+y^2}\)
\(=\dfrac{1}{x-y}-\dfrac{3xy}{x^3-y^3}+\dfrac{x-y}{x^2+xy+y^2}\)
\(=\dfrac{1}{x-y}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{x-y}{x^2+xy+y^2}\) MTC: \(\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{\left(x-y\right)\left(x-y\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{\left(x^2+xy+y^2\right)-3xy+\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x^2-2xy+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(1+4\right)\left(x^2+4y^2\right)\ge\left(x+4y\right)^2\)
\(\Rightarrow5\left(x^2+4y^2\right)\ge\left(x+4y\right)^2\)
\(\Rightarrow5\left(x^2+4y^2\right)\ge\left(x+4y\right)^2=1^2=1\)
\(\Rightarrow5\left(x^2+4y^2\right)\ge1\Rightarrow x^2+4y^2\ge\dfrac{1}{5}\)
Đẳng thức xảy ra khi \(x=y=\dfrac{1}{5}\)
x^2 +4y^2 >= 1/5 ta có x+4y=1 => x=1-4y
=> x^2 +4y^2-1/5 >=0
thay x=1-4y vào ta đk
1-8y+16Y^2 +4y^2 -1/5 >=0
20y^2-8y+4/5>=0
5(2y-2/5)>=0(luôn đúng )
suy ra đpcm