cho x+2y=5. CMR: x^2+y^2>5
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Áp dụng BĐT Bunhiacopxky ta có :
﴾ x + 2y ﴿^2 <= ﴾ 1^2 + 2^2 ﴿﴾ x^2 + y^2 ﴿
5^2 <= 5﴾ x^2 + y^2 ﴿
5﴾ x^2 + y^2 ﴿ >= 25
x^2 + y^2 >= 25/5
x^2 + y^2 >= 5
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Có x2 + y2 - 4x - 2y +5 = ( x2 - 4x + 4) + ( y2 - 2y + 1) = (x-2)2 + (y-1)2
Vì (x-2)2 >= 0 với mọi x, (y-1)2 >=0 với mọi y
=> (x-2) + (y-1) >=0 với mọi x,y hay x2 + y2 - 4x - 2y +5 >=0 (đpcm)
\(x^2+y^2-4x-2y+5=\left(x^2-4x+4\right)+\left(y^2-2y+1\right)\)
\(=\left(x-2\right)^2+\left(y-1\right)^2\ge0\)
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(1)
(x+1)(x-7)+17>0
<=>x^2-6x+9+1>0
<=>(x-3)^2+1>0(dpcm)
..
(7)
-y^2+4y-4-|x+1|≤0
<=>-(y-2)^2-|x+1|≤0
sum 2 so khong duong ko the la so (+)=>dpcm
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Theo BĐT Cauchy cho 2 số dương, ta có:
\(2x^2+y^2+5=\left(x^2+y^2\right)+\left(x^2+1\right)+4\ge2\left(xy+x+2\right)\)
\(\Rightarrow\frac{x}{2x^2+y^2+5}\le\frac{x}{2\left(xy+x+2\right)}\)(1)
Tương tự ta có: \(\frac{2y}{6y^2+z^2+6}\le\frac{2y}{4\left(yz+y+1\right)}=\frac{y}{2\left(yz+y+1\right)}\)(2)
\(\frac{4z}{3z^2+4x^2+16}\le\frac{4z}{4\left(zx+2z+2\right)}=\frac{z}{zx+2z+2}\)(3)
Cộng theo vế của 3 BĐT (1), (2), (3), ta được: \(\frac{x}{2x^2+y^2+5}+\frac{2y}{6y^2+z^2+6}+\frac{4z}{3z^2+4x^2+16}\)
\(\le\frac{1}{2}\left(\frac{x}{xy+x+2}+\frac{y}{yz+y+1}+\frac{2z}{zx+2z+2}\right)\)
\(=\frac{1}{2}\left(\frac{zx}{xyz+xz+2z}+\frac{xyz}{xyz^2+xyz+xz}+\frac{2z}{zx+2z+2}\right)\)
\(=\frac{1}{2}\left(\frac{zx}{2+xz+2z}+\frac{2}{2z+2+xz}+\frac{2z}{zx+2z+2}\right)\)(Do xyz = 2)
\(=\frac{1}{2}.\frac{zx+2z+2}{zx+2z+2}=\frac{1}{2}\)
Đẳng thức xảy ra khi x = y = 1; z = 2
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1.(x+1)(x-7)+17=(x-3)2+1>0
2.-20-(x-5)(x+3)=-34-(x-1)2<0
3.-2(x+3)-(x-2)(x+2)=-(x+1)2-1<0
4.x2+y2+2x+2y+3=(x+1)2+(y+1)2+1>0
5.2x2+2x+y2+2y+5=2(x+1/2)2+(y+1)2+2>0
6.2x2+2y2+2xy+2x+4y+6=(x+y)2+(x+1)2+(y+2)2+1>0
7.-y2+4y-4-/x+1/=-(y-2)2-/x+1/≤0
Áp dụng BĐT Bunhiacopxky ta có :
( x + 2y )2 <= ( 12 + 22 )( x2 + y2 )
52 <= 5( x2 + y2 )
5( x2 + y2 ) >= 25
x2 + y2 >= 25/5
x2 + y2 >= 5