cho x , y , z > 0 thỏa mãn xy + yz + zx = 3xyz
CMR: \(A=\frac{x^3}{z+x^2}+\frac{y^3}{x+y^2}+\frac{z^3}{y+z^2}\ge\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
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Chứng minh rằng:a, nếu x+y1 thì frac{x}{y^3-1}+frac{y}{x^3-1}+frac{2left(xy-2right)}{x^2y^2+3}0 b, nếu x,y,z khác -1 thìfrac{xy+2x+1}{xy+x+y+1}+frac{yz+2y+1}{yz+z+y+1}+frac{zx+2z+1}{zx+z+x+1}3c, Cho x,y,z đôi một khác nhau thỏa mãnfrac{x}{y-z}+frac{y}{z-x}+frac{z}{x-y}0 thìfrac{x}{left(y-zright)^2}+frac{y}{left(z-xright)^2}+frac{z}{left(x-yright)^2}0
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Chứng minh rằng:
a, nếu x+y=1 thì \(\frac{x}{y^3-1}+\frac{y}{x^3-1}+\frac{2\left(xy-2\right)}{x^2y^2+3}=0\)
b, nếu x,y,z khác -1 thì\(\frac{xy+2x+1}{xy+x+y+1}+\frac{yz+2y+1}{yz+z+y+1}+\frac{zx+2z+1}{zx+z+x+1}=3\)
c, Cho x,y,z đôi một khác nhau thỏa mãn\(\frac{x}{y-z}+\frac{y}{z-x}+\frac{z}{x-y}=0\) thì\(\frac{x}{\left(y-z\right)^2}+\frac{y}{\left(z-x\right)^2}+\frac{z}{\left(x-y\right)^2}=0\)
Cho các sô thực dương x,y,z thỏa mãn xy+yz+zx=3 .CMR:\(\frac{1}{xyz}+\frac{4}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\ge\frac{3}{2}\)
cho x,y,z là các số dương thỏa mãn \(xyz=\frac{1}{2}\)CMR : \(\frac{yz}{x^2\left(y+z\right)}+\frac{zx}{y^2\left(x+z\right)}+\frac{xy}{z^2\left(y+x\right)}\ge xy+yz+zx\)
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 các số dương x,y,z thỏa mãn: xy + yz + zx = 3xyz. Chứng minh rằng
\(\frac{x^3}{x^2+z}+\frac{y^3}{y^2+x}+\frac{z^3}{z^2+y}\ge\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Theo GT : \(xy+yz+xz=3xyz\Rightarrow\frac{xy+yz+xz}{xyz}=3\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)
\(\frac{x^3}{x^2+z}=\frac{x\left(x^2+z\right)}{x^2+z}-\frac{xz}{x^2+z}=x-\frac{xz}{x^2+z}\ge x-\frac{xz}{2x\sqrt{z}}=x-\frac{\sqrt{z}}{2}\)
Tương tự , ta có : \(\frac{y^3}{y^2+x}\ge y-\frac{\sqrt{x}}{2}\) ; \(\frac{z^3}{z^2+y}\ge z-\frac{\sqrt{y}}{2}\)
\(\Rightarrow\frac{x^3}{x^2+z}+\frac{y^3}{y^2+z}+\frac{z^3}{z^2+y}\ge x+y+z-\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{2}\)
Vì x ; y ; z dương , áp dụng BĐT Cô - si , ta có :
\(x+1\ge2\sqrt{x};y+1\ge2\sqrt{y};z+1\ge2\sqrt{z}\)
\(\Rightarrow x+y+z+3\ge2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)
=> \(\frac{x+y+z+3}{2}\ge\sqrt{x}+\sqrt{y}+\sqrt{z}\) => BĐT được c/m
Tiếp tục AD BĐT Cô - si , ta có :
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{\frac{1}{xyz}}=9\)
\(\Rightarrow x+y+z\ge\frac{9}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=\frac{9}{3}=3\) => BĐT được c/m
Có : \(\frac{x^3}{x^2+z}+\frac{y^3}{y^2+x}+\frac{z^3}{z^2+y}\ge x+y+z-\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{2}\ge x+y+z-\frac{x+y+z+3}{4}=\frac{3x+3y+3z-3}{2}\ge\frac{3.3-3}{4}=\frac{3}{2}=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Dấu " = " xảy ra \(\Leftrightarrow x=y=z=1\)
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Cho x,y,z là các số thực dương thỏa mãn xy+yz+zx=1
Chứng minh rằng
\(\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+z^2}\ge\frac{2}{3}\left(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\right)^3\)
Ta có \(1+x^2=x^2+xy+yz+xz=\left(x+z\right)\left(x+y\right)\)
Khi đó BĐT <=>
\(\frac{1}{\left(x+y\right)\left(x+z\right)}+\frac{1}{\left(y+z\right)\left(x+z\right)}+\frac{1}{\left(x+y\right)\left(y+z\right)}\ge\frac{2}{3}\left(\frac{x}{\sqrt{\left(x+z\right)\left(x+y\right)}}+...\right)\)
<=> \(\frac{x+y+z}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\frac{x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}}{\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}}\right)^3\)
<=>\(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge\frac{1}{3}\left(x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}\right)^3\)
<=> \(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\right)^3\)(1)
Xét \(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge\frac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)
<=> \(9\left[xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\right]\ge8\left(xy\left(x+y\right)+xz\left(x+z\right)+yz\left(y+z\right)+3xyz\right)\)
<=> \(xy\left(y+x\right)+yz\left(y+z\right)+xz\left(x+z\right)\ge6xyz\)
<=> \(x\left(y-z\right)^2+z\left(x-y\right)^2+y\left(x-z\right)^2\ge0\)luôn đúng
Khi đó (1) <=>
\(\left(x+y+z\right).\frac{2\sqrt{2}}{3}.\sqrt{x+y+z}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+....\right)^3\)
<=> \(\sqrt{2\left(x+y+z\right)}\ge\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\)
Áp dụng buniacopxki cho vế phải ta có
\(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\le\sqrt{\left(x+y+z\right)\left(3-xy-yz-xz\right)}\)
\(=\sqrt{2\left(x+y+z\right)}\)
=> BĐT được CM
Dấu bằng xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
cho x;y;z là các số thực dương thỏa mãn x+y+z=1.CMR:
\(\frac{xy}{x^2+y^2}+\frac{yz}{y^2+z^2}+\frac{zx}{z^2+x^2}+\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\frac{15}{4}\)
vì x+y+z=1nên
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\)\(\frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{z}\)\(=3+\left(\frac{x}{y}+\frac{y}{z}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)\)=\(3+\frac{x^2+y^2}{xy}+\frac{y^2+z^2}{yz}+\frac{x^2+z^2}{xz}\)
nen \(\frac{xy}{x^2+y^2}+\frac{yz}{y^2+z^2}+\frac{xz}{x^2+z^2}+\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\) =\(\left(\frac{xy}{x^2+y^2}+\frac{x^2+y^2}{4xy}\right)+\left(\frac{yz}{y^2+z^2}+\frac{y^2+z^2}{4yz}\right)+\left(\frac{xz}{x^2+z^2}+\frac{x^2+z^2}{xz}\right)+\frac{3}{4}\)
\(\ge2.\frac{1}{2}+\frac{2.1}{2}+\frac{2.1}{2}+\frac{3}{4}=\frac{15}{4}\)(dpcm)
dau = xay ra khi x=y=z=1/3
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cho 3 số thực dương x;y;z thỏa mãn x+y+z<=3/2. tìm GTNN của biểu thức:
\(p=\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}\)
Áp dụng bđt AM-GM ta có
\(P\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2.\left(yz+1\right)^2.\left(zx+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}=A\)
Ta có \(A=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{zx+1}{z}\right)}=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Áp dụng bđt AM-GM ta có
\(A\ge3\sqrt[3]{8\sqrt{\frac{xyz}{xyz}}}=3.2=6\)
\(\Rightarrow P\ge6\)
Dấu "=" xảy ra khi x=y=z=\(\frac{1}{2}\)
Làm tiếp bài ღ๖ۣۜLinh's ๖ۣۜLinh'sღ] ★we are one★ chớ hình như bị ngược dấu ó.Do mình gà nên chỉ biết cô si mù mịt thôi ạ
\(3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
\(=3\sqrt[3]{\left(y+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}\right)\left(z+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4y}\right)\left(x+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4z}\right)}\)
\(\ge3\sqrt[3]{5\sqrt[5]{\frac{y}{256x^4}}\cdot5\sqrt[5]{\frac{z}{256y^4}}\cdot5\sqrt[5]{\frac{x}{256z^4}}}\)
\(=3\sqrt[3]{125\sqrt[5]{\frac{xyz}{256^3\left(xyz\right)^4}}}\)
\(=15\sqrt[3]{\sqrt[5]{\frac{1}{256^3\left(xyz\right)^3}}}\)
\(\ge15\sqrt[15]{\frac{1}{256^3\cdot\left(\frac{x+y+z}{3}\right)^9}}\)
\(\ge15\sqrt[15]{\frac{1}{256^3\cdot\frac{1}{2^9}}}=\frac{15}{2}\)
Dấu "=" xảy ra tại \(x=y=z=\frac{1}{2}\)
Không phải ngược đâu nha mọi người,dấu bằng không xảy ra nhé!
cho 3 số thực dương x;y;z thỏa mãn x+y+z<=3/2. tìm GTNN của biểu thức :
\(P=\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}\)
Bài này thì AM-GM thôi
\(P=\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}\)
Sử dụng BĐT AM-GM cho 3 số không âm ta có :
\(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)^2}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(zx+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}\)
\(=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{zx+1}{z}\right)}\)
\(=3\sqrt[3]{\left(\frac{xy}{x}+\frac{1}{x}\right)\left(\frac{yz}{y}+\frac{1}{y}\right)\left(\frac{zx}{z}+\frac{1}{z}\right)}=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Tiếp tục sử dụng AM-GM cho 2 số không âm ta được :
\(3\sqrt[3]{\left(2\sqrt[2]{y\frac{1}{x}}\right)\left(2\sqrt[2]{z\frac{1}{y}}\right)\left(2\sqrt[2]{x\frac{1}{z}}\right)}\ge3\sqrt[3]{\left(2\sqrt{\frac{y}{x}}\right)\left(2\sqrt{\frac{z}{y}}\right)\left(2\sqrt{\frac{x}{z}}\right)}\)
\(=3\sqrt[3]{8\left(\sqrt{\frac{y}{x}}.\sqrt{\frac{z}{y}}.\sqrt{\frac{x}{z}}\right)}=3\sqrt[3]{8.\sqrt{\frac{xyz}{xyz}}}=3\sqrt[3]{8}=3.2=6\)
Đẳng thức xảy ra khi và chỉ khi \(x=y=z=\frac{1}{2}\)
Vậy \(Min_P=6\)đạt được khi \(x=y=z=\frac{1}{2}\)