Cho x,y>0 2(x+y+z)+xy+yz+zx=9xyz Cm 1/x^2+1/z^2+1/y^2>=3
Những câu hỏi liên quan
x,y,z>0. CMR \(x^2+y^2+z^2+\frac{9xyz}{x+y+z}\ge2\left(xy+yz+zx\right)\)
Từ bất đẳng thức trên + x+y+z=1 làm sao để suy ra \(9xyz\ge4\left(xy+yz+zx\right)-1\)
cho x,y,z là 3 số không âm t/m x+y+z=1. cmr xy+yz+zx<=2/7+ (9xyz)/7
Cho x; y; z >0, thoả mãn: 1/xy+ 1/yz+1/zx =1
Q= x/√yz × (x^2 +1)+ y/√zx × (y^2 +1) + z/√xy × ( z^2 +1)
Cho x,y,z>0 thỏa mãn xy+yz+zx=1. Chứng minh \(\frac{x}{x^2-yz+3}+\frac{y}{y^2-zx+3}+\frac{z}{z^2-xy+3}\ge\frac{1}{x+y+z}\)
cho 3 số dương 0<x<y<z<1.CM X/YZ+1 + Y/ZX+1 + Z/XY+1<2
Áp dụng BĐT Cauchy:[TEX]xyzgeq (x+y-z)(y+z-x)(x+z-y)(6-2x)(6-2y)(6-2z) 216-72(x+y+z)+24(xy+yz+zx)-8xyz24(xy+yz+xz)-8xyz-216 Rightarrow 9xyzgeq 24(xy+yz+xz)-216 Rightarrow xyzgeq frac{8}{3}(xy+yz+xz)-24 Rightarrow x^{2}+y^2+z^2-xy-yz-zx+xyzgeq x^{2}+y^2+z^2+frac{5}{3}(xy+yz+zx)-24 Leftrightarrow (x+y+z)^{2}-frac{1}{3}( xy+yz+zx)-24geq (x+y+z)^{2}-24-frac{1}{9}(x+y+z)^{2}8[/TEX]Dấu xảy ra khi [TEX]xyz2[/TEX]
Đọc tiếp
Áp dụng BĐT Cauchy:
[TEX]xyz\geq (x+y-z)(y+z-x)(x+z-y)=(6-2x)(6-2y)(6-2z) \\ =216-72(x+y+z)+24(xy+yz+zx)-8xyz=24(xy+yz+xz)-8xyz-216 \\ \Rightarrow 9xyz\geq 24(xy+yz+xz)-216 \\ \Rightarrow xyz\geq \frac{8}{3}(xy+yz+xz)-24 \\ \Rightarrow x^{2}+y^2+z^2-xy-yz-zx+xyz\geq x^{2}+y^2+z^2+\frac{5}{3}(xy+yz+zx)-24 \\ \Leftrightarrow (x+y+z)^{2}-\frac{1}{3}( xy+yz+zx)-24\geq (x+y+z)^{2}-24-\frac{1}{9}(x+y+z)^{2}=8[/TEX]
Dấu "=" xảy ra khi [TEX]x=y=z=2[/TEX]
cho 3 số x ,y ,z #0 thõa mãn 1/x + 1/y +1/z=0 . tính : P =(xy/z^2 + yz/x^2 +zx/y^2 -2)^2013
cho Q= \(\sqrt{x^2-xy+y^2}\)+ \(\sqrt{y^2-yz+z^2}\)+\(\sqrt{z^2-zx+x^2}\) với x,y,z > 0 x+y+z=3
CM : Q ≥ 3
\(x^2-xy+y^2=\dfrac{1}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2\ge\dfrac{1}{4}\left(x+y\right)^2\)
\(\Rightarrow\sqrt{x^2-xy+y^2}\ge\sqrt{\dfrac{1}{4}\left(x+y\right)^2}=\dfrac{1}{2}\left(x+y\right)\)
Tương tự: \(\sqrt{y^2-yz+z^2}\ge\dfrac{1}{2}\left(y+z\right)\); \(\sqrt{z^2-zx+x^2}\ge\dfrac{1}{2}\left(z+x\right)\)
Cộng vế:
\(Q\ge\dfrac{1}{2}\left(x+y\right)+\dfrac{1}{2}\left(y+z\right)+\dfrac{1}{2}\left(z+x\right)=x+y+z=3\) (đpcm)
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Cho x,y,z > 0 thỏa mãn xy + yz +zx = 1.Chứng minh
\(\frac{x-y}{z^2+1}\)+\(\frac{y-z}{x^2+1}\)+\(\frac{z-x}{y^2+1}\)=0
\(\dfrac{x-y}{z^2+1}=\dfrac{x-y}{z^2+xy+yz+zx}=\dfrac{x-y}{z\left(z+y\right)+x\left(z+y\right)}=\dfrac{x-y}{\left(x+z\right)\left(z+y\right)}\)
Tương tự: \(\dfrac{y-z}{x^2+1}=\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}\);\(\dfrac{z-x}{y^2+1}=\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)
Cộng vế với vế \(\Rightarrow VT=\dfrac{x-y}{\left(x+z\right)\left(y+z\right)}+\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}+\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)+\left(y-z\right)\left(y+z\right)+\left(z-x\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\dfrac{x^2-y^2+y^2-z^2+z^2-x^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=0\)(đpcm)
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Cho 3 số x,y vả z thoả mãn 1/x+1/y+1/z=0. Hãy tính A= yz/x^2+zx/y^2+xy/z^2
CÓ:\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)
\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)=-\frac{1}{z^3}\)
\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}-\frac{3}{xyz}=-\frac{1}{z^3}\)
\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)
\(A=\frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz\cdot\frac{3}{xyz}=3\)
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