Question

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

với 0<=x,y,z<=1

chứng minh P>=2

Answer 1

Lời giải:

Vì \(x,y,z\in [0;1]\Rightarrow xy; yz,xz\geq xyz\)

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

\(x,y,z\in [0;1]\Rightarrow \left\{\begin{matrix} (x-1)(y-1)\geq 0\\ (xy-1)(z-1)\geq 0\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} xy+1\geq x+y\\ xyz+1\geq xy+z\end{matrix}\right.\)

\(\Rightarrow xyz+2+xy\geq x+y+z+xy\)

\(\Leftrightarrow xyz+2\geq x+y+z\)

Mà: \(xyz+2\leq 2xyz+2=2(xyz+1)\)

\(\Rightarrow x+y+z\leq 2(xyz+1)(**)\)

Từ \((*); (**)\Rightarrow P\leq \frac{2(xyz+1)}{xyz+1}=2\) (đpcm)

Dấu "=" xảy ra khi \((x,y,z)=(1,1,0)\)