1. Không mất tính tổng quát, giả sử: \(x\ge y\ge z\)
\(\Rightarrow x\left(y-z\right)\left(y-x\right)\le0\Leftrightarrow x\left(y^2-xy-yz+xz\right)\le0\)
\(\Leftrightarrow xy^2+yz^2+zx^2\le x^2y+yz^2+xyz\)
Ta chứng minh
\(x^2y+yz^2\le2\Leftrightarrow y\left(x^2+z^2\right)\le2\Leftrightarrow y\left(3-y^2\right)\le2\)
\(\Leftrightarrow y^3+2\ge3y\)
Áp dụng bđt cô - si cho 3 số không âm:
\(y^3+2=y^3+1+1\ge3\sqrt[3]{y^3.1.1}=3y\)
Lúc đó \(\Leftrightarrow xy^2+yz^2+zx^2\le2+xyz\)
Đẳng thức xảy ra khi x = y = z = 1
2. Ta có:
\(P=\text{Σ}_{cyc}\frac{x}{2+y}=\frac{\text{Σ}_{cyc}\left[x\left(z+2\right)\left(x+2\right)\right]}{\left(z+2\right)\left(y+2\right)\left(x+2\right)}\)
\(=\frac{4\left(\text{Σ}_{cyc}a\right)+2\left(\text{Σ}_{cyc}x^2\right)+2\left(\text{Σ}_{cyc}xy\right)+\text{Σ}_{cyc}\left(x^2z\right)}{4\left(x+y+z\right)+8+xyz+3\left(xy+yz+zx\right)}\)
Theo phần 1 thì \(xy^2+yz^2+zx^2\le2+xyz\)
\(\Rightarrow P\le\frac{4\left(x+y+z\right)+8+xyz+2\left(xy+yz+zx\right)}{4\left(x+y+z\right)+8+xyz+2\left(xy+yz+zx\right)}=1\)
Dấu "=" khi x = y = z = 1