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

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

sửa lại cái dòng thứ 2 từ dưới lên nha

vt\(\le2020\left(x+y+z\right)=2020.2019xyz=\) vp (dpcm)

bằng .vp chư ko phải = .vt nha

\(\hept{\begin{cases}x,y,z>0\\x+y+z=xyz\end{cases}}\)

\(\Rightarrow\frac{1}{xy} +\frac{1}{yz}+\frac{1}{zx}=1\)

Có : \(\frac{1}{\sqrt{1+x^2}}=\frac{1}{\sqrt{\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+x^2}}\le\frac{1}{2.\sqrt{\frac{x^2y}{xyz}}}\le\frac{1}{2}\)

\(\frac{1}{\sqrt{1+y^2}}=\frac{1}{\sqrt{\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+y^2}}\le\frac{1}{2\sqrt{\frac{y^2z}{xyz}}}\le\frac{1}{2}\)

\(\frac{1}{\sqrt{1+z^2}}=\frac{1}{\sqrt{\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+z^2}}\le\frac{1}{2\sqrt{\frac{z^2x}{xyz}}}\le\frac{1}{2}\)

\(\Rightarrow\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}+\frac{1}{\sqrt{1+z^2}}\le\frac{3}{2}\)

Vậy P max = 3/2

\(VT=\sum\frac{x}{\sqrt{1+x^2}}=\sum\frac{x}{\sqrt{xy+xz+yz+x^2}}=\sum\frac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}\le\frac{1}{2}\sum\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\)\(\Rightarrow VT\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{y+z}+\frac{y}{x+y}+\frac{z}{x+z}+\frac{z}{y+z}\right)=\frac{3}{2}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)

Ta co: \(1+x^2=xy+yz+zx+x^2=\left(x+y\right)\left(x+z\right)\)

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

Thê vào ta được

\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\left(xy+yz+zx\right)=2\)

ta có :

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

tương tự ta sẽ có :

\(A=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+yz+xz\right)=2\)

helloo

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 \(x=y=z=\frac{1}{\sqrt{3}}\)

Vì xyz=1\(\Rightarrow x^2\left(y+z\right)\ge2x^2\sqrt{yz}=2x\sqrt{x}\)

Tương tự \(y^2\left(z+x\right)\ge2y\sqrt{y};z^2=\left(x+y\right)\ge2z\sqrt{z}\)

\(\Rightarrow P\ge\frac{2x\sqrt{x}}{y\sqrt{y}+2z\sqrt{z}}+\frac{2y\sqrt{y}}{z\sqrt{z}+2x\sqrt{x}}+\frac{2z\sqrt{z}}{x\sqrt{x}+2y\sqrt{y}}\)

Đặt \(x\sqrt{x}+2y\sqrt{y}=a;y\sqrt{y}+2z\sqrt{z}=b;z\sqrt{z}+2x\sqrt{x}=c\)

\(\Rightarrow x\sqrt{x}=\frac{4c+a-2b}{9};y\sqrt{y}=\frac{4a+b-2c}{9};z\sqrt{z}=\frac{4b+c-2a}{9}\)

\(\Rightarrow P\ge\frac{2}{9}\left(\frac{4c+a-2b}{b}+\frac{4a+b-2c}{a}+\frac{4b+c-2a}{b}\right)\)

\(=\frac{2}{9}\text{ }\left[4\left(\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\right)+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-6\right]\ge\frac{2}{9}\left(4.3+2-6\right)=2\)

Min P =2 khi và chỉ khi a=b=c khi va chỉ khi x=y=z=1

mình cũng định hỏi câu này sorry mình cx chẳng bt

Tui cũng ko bt, tui đang học lớp 6

tui không biết tôi học lớp 5