Ta có:

\(xy+yz+zx=4xyz\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\)

\(P=\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\)

\(\le\frac{1}{16}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}+\frac{2}{y}+\frac{1}{z}+\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\right)\)

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

áp dụng cô si sháp cho 4 số ta được :

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{16}{a+b+c+d}\)  Luôn đúng , ( tự chứng minh )

\(\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\ge\frac{1}{a+b+c+d}\) luôn luôn đúng

áp dụng vào  P ta được như sau

\(\frac{1}{x+x+y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\) luôn đúng :))

\(\frac{1}{x+y+y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}\right)\)

\(\frac{1}{x+y+z+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}\right)\)

Cộng tất cả vào ta được

\(P\le\frac{1}{16}\left(\frac{4}{x}+\frac{4}{y}+\frac{4}{z}\right)\Leftrightarrow P\le\frac{1}{4}\left(x+y+z\right)\)

Thèo đề \(xy+yz+xz=4xyz\Leftrightarrow xy+yz+xz=xyz+xyz+xyz+xyz\)

Tao cũng éo hiểu tại sao nó = nhau được

1 đề sai  , 2 tao sai thế thôi

? cho a,b,c tìm x,y,z là seo?

chắc đề cho x+y+z=1

\(=>\sqrt{x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\)

\(=>\dfrac{x}{x+\sqrt{\left(x+y\right)\left(x+z\right)}}\le\dfrac{x}{x+\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}}\)

\(=\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

làm tương tự với \(\dfrac{y}{y+\sqrt{y+xz}},\dfrac{z}{z+\sqrt{z+xy}}\)

\(=>A\le\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\) dấu"=" xảy ra<=>x=y=z=`/3

\(xy+yz+zx=3xyz\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3\)

Có \(\dfrac{1}{x+2y+3z}=\dfrac{1}{\left(x+y\right)+\left(y+z\right)+2z}\le\dfrac{1}{9}\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{2z}\right)\le\dfrac{1}{9}\left(\dfrac{1}{4x}+\dfrac{1}{4y}+\dfrac{1}{4y}+\dfrac{1}{4z}+\dfrac{1}{2z}\right)=\dfrac{1}{9}\left(\dfrac{1}{4x}+\dfrac{1}{2y}+\dfrac{3}{4z}\right)\)

Tương tự cx có: \(\dfrac{1}{y+2z+3x}\le\dfrac{1}{9}\left(\dfrac{1}{4y}+\dfrac{1}{2z}+\dfrac{3}{4x}\right)\);\(\dfrac{1}{z+2x+3y}\le\dfrac{1}{9}\left(\dfrac{1}{4z}+\dfrac{1}{2x}+\dfrac{3}{4y}\right)\)

Cộng vế với vế \(\Rightarrow\Sigma\dfrac{1}{x+2y+3z}\le\dfrac{1}{9}\left(\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{3}{4}\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{2}\)

Dấu "=" xayra khi x=y=z=1

Vậy \(P_{max}=\dfrac{1}{2}\)

Khó thế, mới lớp 8, làm mãi ko ra

\(xy+yz+zx=8xyz\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=8\)

\(\Rightarrow\dfrac{8}{x}+\dfrac{8}{y}+\dfrac{8}{z}=64\)

Ta có: \(\dfrac{8}{x}+\dfrac{8}{y}+\dfrac{8}{z}\)

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

(sau dấu chấm là bốn số tương tự).

\(\ge^{Cauchy-Schwarz}\dfrac{8^2}{6x+y+z}+\dfrac{8^2}{6y+z+x}+\dfrac{8^2}{6z+x+y}\)

\(\Rightarrow64\ge\dfrac{8^2}{6x+y+z}+\dfrac{8^2}{6y+z+x}+\dfrac{8^2}{6z+x+y}\)

\(\Rightarrow\dfrac{1}{6x+y+z}+\dfrac{1}{6y+z+x}+\dfrac{1}{6z+x+y}\le1\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{3}{8}\)

Vậy \(Max\) của biểu thức đã cho là 1.

BĐT tương đương:

\(\frac{1}{z\left(1+\frac{1}{x}\right)}+\frac{1}{x\left(1+\frac{1}{y}\right)}+\frac{1}{y\left(1+\frac{1}{z}\right)}\ge2\)

Từ giả thiết:

\(xy+yz+zx+2xyz=1\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+2=\frac{1}{xyz}\)

Đặt \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)=\left(a;b;c\right)\Rightarrow a+b+c+2=abc\)

\(\Rightarrow a+b+c+2\le\frac{1}{27}\left(a+b+c\right)^3\)

\(\Leftrightarrow\left(a+b+c\right)^3-27\left(a+b+c\right)-54\ge0\)

\(\Leftrightarrow\left(a+b+c-6\right)\left(a+b+c+3\right)^2\ge0\)

\(\Leftrightarrow a+b+c\ge6\)

BĐT trở thành: \(\frac{c}{1+a}+\frac{a}{1+b}+\frac{b}{1+c}\ge2\)

Thật vậy, ta có:

\(VT=\frac{a^2}{a+ab}+\frac{b^2}{b+bc}+\frac{c^2}{c+ca}\ge\frac{\left(a+b+c\right)^2}{a+b+c+ab+bc+ca}\ge\frac{3\left(a+b+c\right)^2}{3\left(a+b+c\right)+\left(a+b+c\right)^2}\)

\(VT\ge\frac{3\left(a+b+c\right)}{3+a+b+c}=\frac{2\left(a+b+c\right)+a+b+c}{a+b+c+3}\ge\frac{2\left(a+b+c\right)+6}{a+b+c+3}=2\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=2\) hay \(x=y=z=\frac{1}{2}\)