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Bạn ghi đề sai ở dữ kiện \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=0\)
Vì điều đó tương đương với \(x=y=z=0\)
Ta có : \(\frac{x^3}{z+x^2}=\frac{x^3+xz-xz}{z+x^2}=x-\frac{xz}{z+x^2}\ge x-\frac{xz}{2x\sqrt{z}}=x-\frac{\sqrt{z}}{2}\ge x-\frac{z+1}{4}\) (Cosi)
Tương tự \(\hept{\begin{cases}\frac{y^3}{x+y^2}\ge y-\frac{x+1}{4}\\\frac{z^3}{y+z^2}\ge z-\frac{y+1}{4}\end{cases}}\)
\(\Rightarrow\frac{x^3}{z+x^2}+\frac{y^3}{x+y^2}+\frac{z^3}{y+z^2}\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\)
Mà \(xy+yz+xz=3xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\Rightarrow x+y+z\ge3\)
\(\Rightarrow\frac{x^3}{z+x^2}+\frac{y^3}{x+y^2}+\frac{z^3}{y+z^2}\ge\frac{9}{4}-\frac{3}{4}=\frac{3}{2}\ge\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
bước cuối sai \(\frac{3}{2}\ge\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\) trong khi \(3\le x+y+z\) ?? :D
+ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\Rightarrow\frac{1}{z}=2-\frac{1}{x}-\frac{1}{y}\)
\(\Rightarrow\frac{1}{z^2}=\left(2-\frac{1}{x}-\frac{1}{y}\right)^2\)
+ \(\frac{2}{xy}-\frac{1}{z^2}=4\Rightarrow\frac{2}{xy}-\left(2-\frac{1}{x}-\frac{1}{y}\right)^2=4\)
\(\Rightarrow\frac{2}{xy}-\left(4+\frac{1}{x^2}+\frac{1}{y^2}-\frac{4}{x}-\frac{4}{y}+\frac{2}{xy}\right)=4\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}-\frac{4}{x}-\frac{4}{y}+8=0\)
\(\Rightarrow\left(\frac{1}{x}-2\right)^2+\left(\frac{1}{y}-2\right)^2=0\) \(\Rightarrow\left\{{}\begin{matrix}\left(\frac{1}{x}-2\right)^2=0\\\left(\frac{1}{y}-2\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}=2\\\frac{1}{y}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{2}\\\frac{1}{z}=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{2}\\z=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow P=\left(\frac{1}{2}+1-\frac{1}{2}\right)^{2018}=1\)
\(\frac{1}{x}+\frac{1}{y}=2-\frac{1}{z}\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}=4+\frac{1}{z^2}-\frac{4}{z}\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}=-\frac{4}{z}\) \(\Rightarrow\frac{1}{z}=-\frac{1}{4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}-\frac{1}{4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)=2\Rightarrow\frac{1}{4x^2}-\frac{1}{x}+1+\frac{1}{4y^2}-\frac{1}{y}+1=0\)
\(\Rightarrow\left(\frac{1}{2x}-1\right)^2+\left(\frac{1}{2y}-1\right)^2=0\Rightarrow\left\{{}\begin{matrix}\frac{1}{2x}-1=0\\\frac{1}{2y}-1=0\end{matrix}\right.\)
\(\Rightarrow x=y=\frac{1}{2}\Rightarrow\frac{1}{z}=2-\left(\frac{1}{x}+\frac{1}{y}\right)=-2\Rightarrow z=-\frac{1}{2}\)
\(\Rightarrow P=\left(\frac{1}{2}+1-\frac{1}{2}\right)^{2018}=1^{2018}=1\)