\(D\le\dfrac{1}{2}\left(1+\dfrac{x}{1+yz}\right)+\dfrac{1}{2}\left(1+\dfrac{y}{1+zx}\right)+\dfrac{z}{2+2xy}\)
\(=1+\dfrac{x}{2\left(1+yz\right)}+\dfrac{y}{2\left(1+zx\right)}+\dfrac{z}{2\left(1+xy\right)}\)
Do \(0\le x;y;z\le1\)
\(\Rightarrow\left(1-x\right)\left(1-y\right)\ge0\Leftrightarrow xy+1\ge x+y\)
\(\Leftrightarrow2\left(xy+1\right)\ge xy+1+x+y\ge x+y+z\)
\(\Rightarrow\dfrac{z}{2\left(1+xy\right)}\le\dfrac{z}{x+y+z}\)
Tương tự: \(\dfrac{x}{2\left(1+yz\right)}\le\dfrac{x}{x+y+z}\) ; \(\dfrac{y}{2\left(1+zx\right)}\le\dfrac{y}{x+y+z}\)
Cộng vế:
\(P\le1+\dfrac{x}{x+y+z}+\dfrac{y}{x+y+z}+\dfrac{z}{x+y+z}=2\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;1;0\right)\)
\(a+b+c+2=abc\)
\(\Leftrightarrow abc+ab+bc+ca+a+b+c+1=ab+bc+ca+2\left(a+b+c\right)+3\)
\(\Leftrightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)=\left(a+1\right)\left(b+1\right)+\left(b+1\right)\left(c+1\right)+\left(c+1\right)\left(a+1\right)\)
\(\Leftrightarrow\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}=1\)
Ta có:
\(a^2+a^2+4\ge\dfrac{1}{3}\left(a+a+2\right)^2=\dfrac{4}{3}\left(a+1\right)^2\)
\(\Rightarrow\sum\dfrac{a+1}{a^2+2}\le\dfrac{3}{2}\sum\dfrac{a+1}{\left(a+1\right)^2}=\dfrac{3}{2}\sum\dfrac{1}{a+1}=\dfrac{3}{2}\)