ko mik 2k7

uk mk hỏi từ k7 trở lên mà

Ta có đánh giá quen thuộc: \(\left(xy+yz+zx\right)^2\ge3xyz\left(x+y+z\right)=3\left(x+y+z\right)\)(Do xyz = 1)\(\Rightarrow\frac{1}{x+y+z}\ge\frac{3}{\left(xy+yz+zx\right)^2}\)

Như vậy, ta cần chứng minh: \(\frac{3}{\left(xy+yz+zx\right)^2}+\frac{1}{3}\ge\frac{2}{xy+yz+zx}\)

Đặt \(t=\frac{1}{xy+yz+zx}\)thì bất đẳng thức trở thành \(3t^2+\frac{1}{3}\ge2t\Leftrightarrow9t^2+1\ge6t\Leftrightarrow\left(3t-1\right)^2\ge0\)*đúng*

Vậy bất đẳng thức được chứng minh

Đẳng thức xảy ra khi \(\hept{\begin{cases}t=\frac{1}{xy+yz+zx}=\frac{1}{3}\\x=y=z>0,xyz=1\end{cases}}\Leftrightarrow x=y=z=1\)

P = \(\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}-3=y.\frac{y}{x+y}+z.\frac{z}{y+z}+x.\frac{x}{z+x}-3\)

\(=y.\left(\frac{y}{x+y}-1+1\right)+z\left(\frac{z}{y+z}-1+1\right)+x\left(\frac{x}{z+x}-1+1\right)-3\)

\(=y\left(\frac{-x}{x+y}+1\right)+z\left(\frac{-y}{y+z}+1\right)+x\left(\frac{-z}{x+z}+1\right)-3\)

\(=x+y+z-\left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{xz}{z+x}\right)-3\)

Lại có \(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=2017\)

\(\Rightarrow x.\frac{x}{x+y}+y.\frac{y}{y+z}+z.\frac{z}{z+x}=2017\)

=> \(x\left(\frac{x}{x+y}-1+1\right)+y\left(\frac{y}{y+z}-1+1\right)+z\left(\frac{z}{z+x}-1+1\right)=2017\)

=> \(x\left(\frac{-y}{x+y}+1\right)+y\left(\frac{-z}{y+z}+1\right)+z\left(\frac{-x}{x+z}+1\right)=2017\)

=> \(x+y+z-\left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x}\right)=2017\)

Khi đó P = 2017 - 3 = 2014