\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=2019\)
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18 tháng 2 2020

Ta có: \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=2019\)

\(\Rightarrow\frac{x+y+z}{xyz}=2019\)

\(\Rightarrow x+y+z=2019xyz\)

\(\Rightarrow2019x^2=\frac{x^2+xy+xz}{yz}\)

\(\Rightarrow2019x^2+1=\frac{x^2+xy+xz+yz}{yz}=\frac{\left(x+y\right)\left(x+z\right)}{yz}\)

\(=\left(\frac{x}{y}+1\right)\left(\frac{x}{z}+1\right)\)

\(\Rightarrow\sqrt{2019x^2+1}=\sqrt{\left(\frac{x}{y}+1\right)\left(\frac{x}{z}+1\right)}\)\(\le\frac{1}{2}\left(\frac{x}{y}+\frac{x}{z}+2\right)=1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)(cô -si)

\(\Rightarrow\frac{x^2+1+\sqrt{2019x^2+1}}{x}\le\frac{x^2+1+1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)}{x}\)\(=x+\frac{2}{x}+\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)

Tương tự ta có: \(\frac{y^2+1+\sqrt{2019y^2+1}}{y}\le y+\frac{2}{y}+\frac{1}{2}\left(\frac{1}{z}+\frac{1}{x}\right)\)

và \(\frac{z^2+1+\sqrt{2019z^2+1}}{z}\le z+\frac{2}{z}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)

Cộng từng vế của các bđt trên, ta được:

\(\text{Σ}_{cyc}\frac{x^2+1+\sqrt{2019x^2+1}}{x}\le x+y+z+3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

Chứng minh được: \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

\(\Rightarrow3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{3\left(xy+yz+zx\right)}{xyz}=\frac{2019.3\left(xy+yz+zx\right)}{2019xyz}\)

\(\le\frac{2019\left(x+y+z\right)^2}{x+y+z}=2019\left(x+y+z\right)\)

\(\Rightarrow VT\le2020\left(x+y+z\right)=2020.2019xyz\)

Vậy \(\text{Σ}_{cyc}\frac{x^2+1+\sqrt{2019x^2+1}}{x}\le2019.2020xyz\left(đpcm\right)\)

21 tháng 3 2020

Theo bài ra ta có:

\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{z}{xyz}+\frac{x}{xyz}+\frac{y}{xyz}=\frac{x+y+z}{xyz}=2019\)

\(\Rightarrow x+y+z=2019xyz\) 

\(\Rightarrow2019x^2=\frac{x^2+xy+xz}{yz}\)

\(\Rightarrow2019x^2+1=\frac{x^2+xy+xz+yz}{yz}=\frac{\left(x+y\right)\left(x+z\right)}{yz}=\left(\frac{x}{y}+1\right)\left(\frac{x}{z}+1\right)\)

\(\Rightarrow\sqrt{2019x^2+1}=\sqrt{\left(\frac{x}{y}+1\right)\left(\frac{x}{z}+1\right)}\le\frac{1}{2}\left(\frac{x}{y}+\frac{x}{z}+2\right)=1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)(Theo BĐT Cosi)

\(\Rightarrow\frac{x^2+1+\sqrt{2019^2+1}}{x}\le\frac{x+1+1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)}{x}=x+\frac{2}{x}+\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)

Tương tự:

\(\frac{y^2+1+\sqrt{2019y^2+1}}{y}\le y+\frac{2}{y}+\frac{1}{2}\left(\frac{1}{z}+\frac{1}{x}\right)\)

\(\frac{z^2+1+\sqrt{2019z^2+1}}{z}\le z+\frac{2}{z}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)

\(\Rightarrow VT\le x+y+z+3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

Chứng minh được: \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

\(\Rightarrow3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{3\left(xy+yz+zx\right)}{xyz}=\frac{2019\cdot3\left(xy+yz+zx\right)}{2019xyz}\le\frac{2019\left(x+y+z\right)^2}{x+y+z}\)\(=2019\left(x+y+z\right)\)
 

\(\Rightarrow VT\le2020\left(x+y+z\right)=2020\cdot2019xyz=VP\)

=> ĐPCM

21 tháng 3 2020

ta có \(x+y+z=2019xyz=>2019x^2=\frac{x^2+xy+xz}{yz}\)

\(=>2019x^2+1=\frac{x^2+xy+xz+yz}{yz}=\frac{\left(x+y\right)\left(x+z\right)}{yz}=\left(\frac{x}{y}+1\right)\left(\frac{x}{z}+1\right)\)

\(=>\sqrt{2019x^2+1}=\sqrt{\left(\frac{x}{y}+1\right)\left(\frac{x}{z}+1\right)}\le\frac{1}{2}\left(\frac{x}{y}+\frac{x}{z}+2\right)=1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)

(theo BDT cô -si)

\(=>\frac{x^2+1+\sqrt{2019x^2+1}}{x}\le\frac{x^2+1+1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)}{x}=x+\frac{2}{x}+\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)

tương tự \(\frac{y^2+1+\sqrt{2019y^2+1}}{z}\le y+\frac{2}{y}+\frac{1}{2}\left(\frac{1}{z}+\frac{1}{x}\right)\)

\(\frac{z^2+1+\sqrt{2019z^2+1}}{z}\le z+\frac{2}{z}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)

=>.vt\(\le x+y+z+3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

chứng minh được \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

=>\(3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{3\left(xy+yz+zx\right)}{2019xyz}\le\frac{2019\left(x+y+z\right)^2}{x+y+z}=2019\left(x+y+z\right)\)

=>.vt\(\le2020\left(x+y+z\right)=2020.2019xyz=\)vt

=> dpcm

21 tháng 3 2020

Ta có: \(2019xyz=x+y+z\)

=> \(2019xy=\frac{x}{z}+\frac{y}{z}+1>1\)\(2019yz=\frac{y}{x}+\frac{z}{x}+1>1\)\(2019xz=\frac{x}{y}+\frac{z}{y}+1>1\)

Ta  lại có: \(x+y+z=2019xyz\)

=> \(2019x\left(x+y+z\right)=2019^2x^2yz\)

=> \(2019x^2+1=\left(2019^2x^2yz-2019xy\right)-\left(2019xz-1\right)\)

=> \(2019x^2+1=\left(2019xy-1\right)\left(2019xz-1\right)\le\frac{\left(2019xy+2019xz-2\right)^2}{4}\)

=> \(\sqrt{2019x^2+1}\le\frac{2019xy+2019xz-2}{2}\)

Tương tự : \(\sqrt{2019y^2+1}\le\frac{2019xy+2019yz-2}{2}\)

\(\sqrt{2019z^2+1}\le\frac{2019xz+2019yz-2}{2}\)

=> \(\frac{x^2+1+\sqrt{2019x^2+1}}{x}+\frac{y^2+1+\sqrt{2019y^2+1}}{y}+\frac{z^2+1+\sqrt{2019z^2+1}}{z}\)

\(\le\)\(\frac{x^2+1+\frac{2019xy+2019xz-2}{2}}{x}+\frac{y^2+1+\frac{2019xy+2019yz-2}{2}}{y}+\frac{z^2+1+\frac{2019xz+2019yz-2}{2}}{z}\)

\(=\frac{2x^2+2019xy+2019xz}{2x}+\frac{2y^2+2019xy+2019yz}{2y}+\frac{2z^2+2019xz+2019yz}{2z}\)

\(=x+\frac{2019}{2}y+\frac{2019}{2}z+y+\frac{2019}{2}x+\frac{2019}{2}z+z+\frac{2019}{2}x+\frac{2019}{2}y\)

\(=2020\left(x+y+z\right)=2020.2019xyz\)

Vậy có điều cần cm

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x=y=z\\x+y+z=2019xyz\end{cases}}\Leftrightarrow x=y=z=\frac{1}{\sqrt{673}}\)

3 tháng 4 2020

helloo

3 tháng 4 2020

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 \(x=y=z=\frac{1}{\sqrt{3}}\)

18 tháng 7 2020

Ta viết lại bất đẳng thức cần chứng minh thành: \(\frac{1}{\sqrt{xy}-4}+\frac{1}{\sqrt{yz}-4}+\frac{1}{\sqrt{zx}-4}\ge-1\)(*)

Theo BĐT Cauchy, ta có: \(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\le\frac{x+y}{2}+\frac{y+z}{2}+\frac{z+x}{2}=x+y+z\)

Mà ta có: \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=9\Rightarrow x+y+z\le3\)nên \(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\le3\)

Theo BĐT Bunyakovsky dạng phân thức: \(\frac{1}{\sqrt{xy}-4}+\frac{1}{\sqrt{yz}-4}+\frac{1}{\sqrt{zx}-4}\)\(\ge\frac{9}{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}-12}\ge\frac{9}{3-12}=-1\)

Suy ra (*) đúng

Vậy bất đẳng thức được chứng minh

Đẳng thức xảy ra khi x = y = z = 1

18 tháng 7 2020

Ine CTV

dễ thấy \(x,y,z< \sqrt{3}\)\(\Rightarrow\)\(\sqrt{xy}-4< 0\); ... 

cauchy-schwarz chỉ dùng cho mẫu dương nha em, bài này lúc trước anh cũng lam sai, noi trước để đừng lục lại :D

16 tháng 1 2020

\(VT\ge\frac{9}{\Sigma_{cyc}\sqrt{xy+x+y}}\ge\frac{9}{\sqrt{\left(1+1+1\right)\left(2x+2y+2z+xy+yz+zx\right)}}\ge\frac{9}{\sqrt{3\left[6+\frac{\left(x+y+z\right)^2}{3}\right]}}=\sqrt{3}\)