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

Tìm \(n\in N\) để \(3^{2n+1}+2^{4n+1}⋮25\)

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

Tương tự => \(M=\frac{1+2y+yz}{\left(y+1\right)\left(z+1\right)}+\frac{1+2z+zx}{\left(1+x\right)\left(z+1\right)}+\frac{1+2x+xy}{\left(1+x\right)\left(y+1\right)}\)

=> \(M=\frac{\left(1+2y+yz\right)\left(1+x\right)+\left(1+2z+zx\right)\left(1+y\right)+\left(1+2x+xy\right)\left(1+z\right)}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)

=>\(M=\frac{6+3\left(x+y+z\right)+3\left(xy+yz+xz\right)}{2+\left(x+y+z\right)+\left(xy+yz+xz\right)}=3\)

Đặt \(a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}\Rightarrow\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca=1\end{matrix}\right.\)

\(K=\frac{\frac{1}{a}}{\sqrt{\frac{1}{bc}\left(1+\frac{1}{a^2}\right)}}+\frac{\frac{1}{b}}{\sqrt{\frac{1}{ac}\left(1+\frac{1}{b^2}\right)}}+\frac{\frac{1}{c}}{\sqrt{\frac{1}{ab}\left(1+\frac{1}{c^2}\right)}}\) \(=\frac{\frac{1}{a}}{\sqrt{\frac{a^2+1}{a^2bc}}}+\frac{\frac{1}{b}}{\sqrt{\frac{b^2+1}{ab^2c}}}+\frac{\frac{1}{c}}{\sqrt{\frac{c^2+1}{abc^2}}}\)

\(=\sqrt{\frac{bc}{a^2+1}}+\sqrt{\frac{ca}{b^2+1}}+\sqrt{\frac{ab}{c^2+1}}\) \(=\sqrt{\frac{bc}{a^2+ab+bc+ca}}+\sqrt{\frac{ca}{b^2+ab+bc+ca}}+\sqrt{\frac{ab}{c^2+ab+bc+ca}}\)

\(=\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\)

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

Dấu "=" \(\Leftrightarrow a=b=c\Leftrightarrow x=y=z=\sqrt{3}\)

Từ dữ kiện đề bài => x + y + z = xyz

Ta có : 

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

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

Tương tự với hai hạng tử còn lại , suy ra 

\(Q\le\frac{1}{2}\left(\frac{x}{x+z}+\frac{x}{x+y}\right)+\frac{1}{2}\left(\frac{y}{x+y}+\frac{y}{y+z}\right)+\frac{1}{2}\left(\frac{z}{z+x}+\frac{z}{z+y}\right)=\frac{3}{2}\)

Vậy Max = 3/2 <=> x = y = z 

Nguồn : Đinh Đức Hùng 

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

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

\(P\le\frac{1}{4}\left(\frac{xy}{x+z}+\frac{xy}{y+z}+\frac{yz}{x+y}+\frac{yz}{x+z}+\frac{xz}{x+y}+\frac{xz}{y+z}\right)\)

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

\("="\Leftrightarrow x=y=z=\frac{1}{3}\)

Khó thế, mới lớp 8, làm mãi ko ra