\(A=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\).Áp dụng BĐT Cauchy-Schwarz,ta có:

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

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

\(\ge3-\frac{9}{\left(x+y+z\right)+\left(1+1+1\right)}=\frac{3}{4}\)

Dấu "=" xảy ra khi x = y = z = 1/3

Vậy A min = 3/4 khi x=y=z=1/3

Bỏ chữ "Áp dụng bđt Cauchy-Schwarz,ta có:"giùm mình,nãy đánh nhầm ở bài làm trước mà quên xóa đi!

À mà để phải là tìm Max mới đúng chứ nhỉ?

Do đó,bạn sửa dòng: \(\ge3-\frac{9}{\left(x+y+z\right)+\left(1+1+1\right)}=\frac{3}{4}\) đến hết thành:

"\(\le3-\frac{9}{\left(x+y+z\right)+\left(1+1+1\right)}=\frac{3}{4}\)

Dấu "=" xảy ra khi x=y=z=1/3

Vậy A max = 3/4 khi x=y=z=1/3

Ta có: \(\left(\sqrt{x+y}\right)^2=\left(\sqrt{x-z}+\sqrt{y-z}\right)^2\)

\(\Leftrightarrow\)\(x+y=x+y-2z+2\sqrt{\left(x-z\right)\left(y-z\right)}\)

\(\Leftrightarrow2z=2\sqrt{\left(x-z\right)\left(y-z\right)}\)

Theo giả thiết, ta có: 

theo giả thiết, ta có: \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=0\Rightarrow\frac{1}{z}-\frac{1}{x}=\frac{1}{y}\)\(\Rightarrow\frac{x-z}{zx}=\frac{1}{y}\Rightarrow x-z=\frac{zx}{y}\)

Tương tự, ta có: \(y-z=\frac{zy}{x}\)

Do đó: \(2\sqrt{\left(x-z\right)\left(y-z\right)}=2\sqrt{\frac{zx}{y}.\frac{zy}{x}}=2z\) (1)

ta có: \(\left(\sqrt{x+y}\right)^2=\left(\sqrt{x-z}+\sqrt{y-z}\right)^2\)

\(\Leftrightarrow2z=2\sqrt{\left(x-z\right)\left(y-z\right)}\)(2)

Thay (2) vào (1) ta thấy (2) luôn đúng

Suy ra ĐPCM

Vì \(x>0,y>0\Rightarrow\frac{1}{x}>0;\frac{1}{y}>0\)

mà \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=0\Rightarrow\frac{1}{z}=\frac{1}{x}+\frac{1}{y}\Rightarrow\frac{1}{z}>0\Rightarrow z>0\)

Ta có: \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=0\Leftrightarrow yz+zx-xy=0\)

\(\Leftrightarrow-z^2=-z^2+yz+zx-xy=-\left(x-z\right)\left(y-z\right)\)

\(\Leftrightarrow z^2=\left(x-z\right)\left(y-z\right)>0\)

\(\Rightarrow z=\sqrt{\left(x-z\right)\left(y-z\right)}\left(z>0\right)\)

Lại có: \(x+y=x-z+y-z+2z\)

\(=\left(x-z\right)+\left(y-z\right)+2\sqrt{\left(x-z\right)\left(y-z\right)}=\left(\sqrt{x-z}+\sqrt{y-z}\right)^2\)

Suy ra \(\sqrt{x+y}=\sqrt{x-z}+\sqrt{y-z}\) (ĐPCM)

\(Q=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{4}{z+4}\right)\le3-\frac{16}{x+y+z+6}=\frac{1}{3}\)

dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\frac{1}{2};\frac{1}{2};-1\right)\)

max=1/3. ra đc rồi ạ

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left[\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

\(\Leftrightarrow x+y=0\) hoặc \(y+z=0\) hoặc \(z+x=0\)

=> ...............................................

ko khó đâu

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

\(\frac{x}{x+1}=\frac{x+1-1}{x+1}=1-\frac{1}{x+1}\) tương tự với y,z

\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)

=> ta đi tìm GTNN của (..)\(A=\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)

đặt x+1=a;y+1=b;z+1=c nội suy cho đỡ đau đầu a+b+c=4

\(B=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) 

\(a+b+c\ge3\sqrt[3]{abc}\)(*)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{a}.\frac{1}{b}.\frac{1}{c}}\)(*)

(*).(**)\(\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\frac{9}{\left(a+b+c\right)}\)

\(\Rightarrow B\ge\frac{9}{4}\Rightarrow A\ge\frac{9}{4}\Rightarrow P\le3-\frac{9}{4}=\frac{3}{4}\)

DS: \(P_{max}=\frac{3}{4}\) đẳng thức khi a=b=c=> x=y=z=1/3

hay was

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

\(=\frac{x}{x+1}-1+\frac{y}{y+1}-1+\frac{z}{z+1}-1+3\)

\(=-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)+3\le\frac{-9}{x+y+z+3}-3=-\frac{9}{4}-3=-\frac{21}{4}\)

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

Vậy \(P_{max}=-\frac{21}{4}\)khi \(x=y=z=\frac{1}{3}\)

:))

\(x\left(x-z\right)+y\left(y-z\right)=0\)\(\Leftrightarrow\)\(x^2+y^2=z\left(x+y\right)\)

\(\frac{x^3}{z^2+x^2}=x-\frac{z^2x}{z^2+x^2}\ge x-\frac{z^2x}{2zx}=x-\frac{z}{2}\)

\(\frac{y^3}{y^2+z^2}=y-\frac{yz^2}{y^2+z^2}\ge y-\frac{yz^2}{2yz}=y-\frac{z}{2}\)

\(\frac{x^2+y^2+4}{x+y}=\frac{z\left(x+y\right)+4}{x+y}=z-x-y+\frac{4}{x+y}+x+y\ge z-x-y+4\)

Cộng lại ra minP=4, dấu "=" xảy ra khi \(x=y=z=1\)

Sai đề nhá, đáng lẽ \(0\le x,y,z\le1\)

Ta dễ có:
\(1+y+zx\le x^2+xy+xz\Rightarrow\frac{x}{1+y+zx}\ge\frac{x}{x^2+xy+xz}=\frac{1}{x+y+z}\)

Tương tự:

\(\frac{y}{1+z+xy}\ge\frac{1}{x+y+z};\frac{z}{1+z+yz}\ge\frac{1}{x+y+z}\)

\(\Rightarrow\frac{x}{1+y+zx}+\frac{y}{1+z+xy}+\frac{z}{1+z+yz}\ge\frac{3}{x+y+z}\)

Đẳng thức xảy ra tại x=y=z=1