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      tích đi rồi tích lại cho

Theo như câu đưới thì

\(\frac{x^3}{y}+\frac{y^3}{z}+\frac{z^3}{x}\ge x^2+y^2+z^2\ge xy+yz+xz\)(bất đẳng thức cosi)

ta caàn chứng minh bđt 

\(\frac{x}{x+yz}+\frac{y}{y+zx}\ge\frac{x}{x+xz}+\frac{y}{y+yz}=\frac{1}{1+z}+\frac{1}{1+z}=\frac{2}{1+z}\)

tương tự + vào, dùng svác sơ

Ta có \(A=\frac{x^4}{x^3+x^2y+xy^2}+...\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^3+y^3+z^3+xy^2+yz^2+zx^2+x^2y+y^2z+z^2x}\)

=> \(A\ge\frac{\left(x^2+y^2+z^2\right)^2}{\left(x^2+y^2+z^2\right)\left(x+y+z\right)}=\frac{x^2+y^2+z^2}{x+y+z}\ge\frac{x+y+z}{3}\left(ĐPCM\right)\)

dấu = xảy ra <=> x=y=z>=0

Thanks

Lời giải:

Áp dụng BĐT AM-GM:

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

Hoàn toàn tương tự với các phân thức còn lại suy ra:

\(\sum \sqrt{\frac{xy}{xy+z}}\leq \frac{1}{2}\left(\frac{x+z}{x+z}+\frac{y+z}{y+z}+\frac{x+y}{x+y}\right)=\frac{3}{2}\)

Ta có đpcm.

Dấu "=" xảy ra khi $x=y=z=\frac{1}{3}$

\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)

\(=\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)

\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\dfrac{1}{3}.\left(x+y+z\right).\dfrac{1}{3}\left(xy+yz+zx\right)\)

\(=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)

\(\ge\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)}.\left(xy+yz+zx\right)\)

\(=\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)^3}\)

\(\Rightarrow3\left(xy+yz+zx\right)^3\le\left(\dfrac{9}{8}\right)^2\)

\(\Rightarrow\left(xy+yz+zx\right)^3\le\dfrac{27}{64}\)

\(\Rightarrow xy+yz+zx\le\dfrac{3}{4}\)