1, cho 3 số x,y,z thỏa mãn xy+yz+zx=3xzy. CMR:\(x^2+y^2+z^2\ge\frac{1}{x^2}1+\frac{1}{y^2}+\frac{1}{z^2}\)
cho các số dương x;y;z thỏa mãn xy+yz+zx=670
CMR: \(\frac{x}{x^2-yz+2010}+\frac{y}{y^2-zx+2010}+\frac{z}{z^2-xy+2010}\ge\frac{1}{x+y+z}\)
Ta có : \(\frac{x}{x^2-yz+2010}+\frac{y}{y^2-xz+2010}+\frac{z}{z^2-xy+2010}\)
\(=\frac{x^2}{x^3-xyz+2010x}+\frac{y^2}{y^3-xyz+2010y}+\frac{z^2}{z^3-xyz+2010z}\)
\(\ge\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2010\left(x+y+z\right)}=\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+3\left(xy+yz+xz\right)\left(x+y+z\right)}\)
\(=\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3+3xy^2+3x^2y+3x^2z+3xz^2+3y^2z+3yz^2}=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\frac{1}{x+y+z}\)
cho x , y , z > 0 thỏa mãn xy + yz + zx = 3xyz
CMR: \(A=\frac{x^3}{z+x^2}+\frac{y^3}{x+y^2}+\frac{z^3}{y+z^2}\ge\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Đặt \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\)
Theo giả thiết,ta có: \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{cd}=\frac{3}{abc}\)
Nhân hai vế với abc: \(a+b+c=3\) tức là \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)
Lại có:\(3=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{xyz}\)
Ta cần c/m: \(A\ge\frac{3}{2}\)
Do x,y,z > 0 áp dụng BĐT Cô si: \(x^3+y^3+z^3\ge3xyz=xy+yz+zx\)
Áp dụng BĐT Cô si: \(A\ge3\sqrt[3]{\frac{x^3y^3z^3}{\left(z+x^2\right)\left(x+y^2\right)\left(y+z^2\right)}}\)
\(=3xyz.\frac{1}{\sqrt[3]{\left(z+x^2\right)\left(x+y^2\right)\left(y+z^2\right)}}\)\(\ge3xyz.\frac{xy+yz+zx}{\left(x+y+z\right)+\left(x^2+y^2+z^2\right)}\)
\(=\frac{3\left(x^2y^2z+xy^2z^2+x^2yz^2\right)}{\left(x+y+z\right)+\left(x^2+y^2+z^2\right)}\ge\frac{3x^2y^2z^2}{\left(x+y+z\right)+\left(x^2+y^2+z^2\right)}\)
\(=\frac{3x^2y^2z^2}{\left(x+y+z\right)+\left(x+y+z\right)^2-2\left(xy+yz+zx\right)}\)
\(=\frac{3x^2y^2z^2}{\left(x+y+z\right)\left(x+y+z+1\right)-6xyz}\)
\(=\frac{3x^2y^2z^2}{xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left[xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+1\right]-6xyz}\)
\(=\frac{3x^2y^2z^2}{3xyz\left[3xyz+1\right]-6xyz}=\frac{3x^2y^2z^2}{9x^2y^2z^2-3xyz}\)
Đặt \(B=\frac{1}{A}=\frac{9x^2y^2z^2-3xyz}{3x^2y^2z^2}\)
Ta sẽ c/m: \(B\ge\frac{2}{3}\).Thật vậy,ta có:
\(B=\frac{1}{A}=\frac{9x^2y^2z^2-3xyz}{3x^2y^2z^2}=3-\frac{3}{3xyz}\)\(=3-\frac{1}{xyz}\ge0\)
Suy ra \(A\ge0?!?\) có gì đó sai sai.Ai biết chỉ giùm
Nghĩ mãi mới ra -.- Để ý cái số mũ 3 trên tử khó mà dùng trực tiếp Cô-si hoặc Bunhia nên phải tách nó ra
Ta có: \(\frac{x^3}{x^2+z}=\frac{x^3+xz}{x^2+z}-\frac{xz}{x^2+z}=x-\frac{xz}{x^2+z}\)
\(\ge x-\frac{xz}{2x\sqrt{z}}\)(Cô-si)
\(=x-\frac{\sqrt{z}}{2}\)
\(\ge x-\frac{z+1}{4}\)(Dùng bđt \(\sqrt{z}\le\frac{z+1}{2}\))
Tương tự \(\frac{y^3}{y^2+z}\ge y-\frac{x+1}{4}\)
\(\frac{z^3}{z^2+y}\ge z-\frac{y+1}{4}\)
Cộng từng vế của các bđt trên lại được
\(A\ge x+y+z-\frac{x+y+z+3}{4}=\frac{3x+3y+3z-3}{4}\)
\(=\frac{3\left(x+y+z\right)}{4}-\frac{3}{4}\)
Từ điều kiện \(xy+yz+zx=3xyz\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)
Áp dụng bđt \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\left(a,b,c>0\right)\)được
\(3=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)
\(\Rightarrow x+y+z\ge3\)
Quay trở lại với A
\(A\ge\frac{3\left(x+y+z\right)}{4}-\frac{3}{4}\ge\frac{3.3}{4}-\frac{3}{4}=\frac{3}{2}=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)(Do \(3=\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\))
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x=y=z\\xy+yz+zx=3\end{cases}\Leftrightarrow x=y=z=1}\)
Vậy .............
tth làm lạ vậy ? Lí giải hộ chỗ \(3=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{xyz}????\)
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
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=673
CMR: \(\frac{x}{x^2-yz+2019}+\frac{y}{y^2-xz+2019}+\frac{z}{z^2-yx+2019}\ge\frac{1}{x+y+z}\)
Đk: $x\geq \frac{1}{2}$
Pt $\Leftrightarrow 4x^2+3x-7=4(\sqrt{x^3+3x^2}-2)+2(\sqrt{2x-1}-1)$
$\Leftrightarrow +4\frac{(x-1)(x+2)^2}{\sqrt{x^3+3x^2}+2}+4\frac{x-1}{\sqrt{2x-1}+1}-(x-1)(4x+7)=0$
$\Leftrightarrow (x-1)[\frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-(4x+7)]=0$
$\Leftrightarrow x=1\vee \frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-4x-7=0$ $(*)$
Xét hàm số $f(x)=\frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-4x-7,x\in [\frac{1}{2};+\infty )$ thì $f(x)>0,\forall x\in [\frac{1}{2};+\infty )$
$\Rightarrow $ Pt $(*)$ vô nghiệm
Cho x;y;z>0 thỏa mãn xyz=1.CMR \(A=\frac{1}{x+y+z}-\frac{2}{xy+yz+zx}\ge\frac{-1}{3}\)
Ta có
\(x^2y^2+y^2z^2+z^2x^2\ge xyz\left(x+y+z\right)\)
\(=>x^2y^2+y^2z^2+z^2x^2+2\left(xyz\right)\left(x+y+z\right)\ge3xyz\left(x+y+z\right)\)
\(=>\left(xy+yz+zx\right)^2\ge3\left(x+y+z\right)\)
\(=>\frac{1}{\left(x+y+z\right)}\ge\frac{3}{\left(xy+yz+zx\right)^2}\)
\(=>A\ge\frac{3}{\left(xy+yz+zx\right)^2}-\frac{2}{xy+yz+zx}\)
đặt
\(\frac{1}{xy+yz+zx}=t\)
\(=>A\ge3t^2-2t\)
mà \(\left(3t-1\right)^2\ge0=>9t^2-6t+1\ge0=>3t^2-2t+\frac{1}{3}\ge0\Rightarrow3t^2-2t\ge-\frac{1}{3}\)
\(=>A\ge-\frac{1}{3}\)(dpcm)
Dấu = xảy ra khi x=y=z=1
tinh tuoi con gai bang 1/4 tuoi me , tuoi con bang 1/5 tuoi me . tuoi con gai cong voi tuoi cua con trai
la 18 tuoi . hoi me bao nhieu tuoi ?
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 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 các sô thực dương x,y,z thỏa mãn xy+yz+zx=3 .CMR:\(\frac{1}{xyz}+\frac{4}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\ge\frac{3}{2}\)