Em thử làm, sai thì thôi nha!

Ta có: \(x^3+y^3+z^3+2\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\)

Áp dụng BĐT AM-GM và BĐT Nesbitt ta có:

\(VT\ge3\sqrt[3]{\left(xyz\right)^3}+2.\frac{3}{2}\ge3+3=6\)

Đẳng thức xảy ra khi x = y = z = 1.

Vậy.....

Is it right???

\(\Rightarrow\left(x+y+z\right)^2\ge\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2\ge3\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)=\dfrac{3\left(x+y+z\right)}{xyz}\Rightarrow x+y+z\ge\dfrac{3}{xyz}\)

\(x+y+z=\dfrac{x+y+z}{3}+\dfrac{2\left(x+y+z\right)}{3}\ge\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{2}{3}.\dfrac{3}{xyz}\ge\dfrac{1}{3}\left(\dfrac{9}{x+y+z}\right)+\dfrac{2}{xyz}=\dfrac{3}{x+y+z}+\dfrac{2}{xyz}\left(đpcm\right)\)

\(dấu"="xảy\) \(ra\Leftrightarrow x=y=z=1\)

Áp dụng bất đẳng thức AM - GM:

\(A=xyz\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

\(\le\left(\frac{x+y+z}{3}\right)^3.\left(\frac{x+y+y+z+z+x}{3}\right)^3\)

\(=\left(\frac{1}{3}\right)^3.\left(\frac{2}{3}\right)^3=\frac{8}{729}\)

\(Max_A=\frac{8}{729}\Leftrightarrow x=y=z=\frac{1}{3}\)

\(x+z+y=1\Leftrightarrow\left(x+y+z\right)^2=1\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2zx\ge3\left(xy+yz+zx\right)=1\Rightarrow M_{max}=\frac{1}{3}.\text{Dâu "=" xay ra }\Leftrightarrow x=y=z=\frac{1}{3}\)

Đơn giản hơn:

Áp dụng bđt quen thuộc \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)

Ta có: \(M\le\frac{\left(x+y+z\right)^2}{3}=\frac{1}{3}\)

Đẳng thức xảy ra khi x = y = z =1/3