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Cach tuong tu
AM-GM \(2+2yz=x^2+y^2+z^2+2yz=x^2+\left(y+z\right)^2\ge2x\left(y+z\right)\)
\(\Rightarrow1+yz\ge x\left(y+z\right)\Rightarrow x^2+x+yz+1\ge x\left(x+y+z+1\right)\)
\(\Rightarrow\frac{x^2}{x^2+x+yz+1}\le\frac{x}{x+y+z+1}\). Se cm \(x+y+z-xyz\le2\), that vay ap dung C-S
\(x+y+z-xyz=x\left(1-yz\right)+\left(y+z\right)\)\(\le\sqrt{\left[x^2+\left(y+z\right)^2\right]\left[\left(1-yz\right)^2+1\right]}\)
\(=\sqrt{2\left(1+yz\right)\left[\left(yz\right)^2-2yz+2\right]}=\sqrt{y^2z^2\left(yz-1\right)+4}\le2\)
\(\Rightarrow M\le\frac{x}{x+y+z+1}+\frac{y+z}{x+y+z+1}+\frac{1}{x+y+z+1}=1\)
Dau "=" xay ra khi x=y=1; z=0
b) C/m: \(\Delta ABC\sim\Delta DAC\left(g.g\right)\Rightarrow AC^2=DC.BC\left(1\right)\)
\(\Delta ABC\sim\Delta DBA\left(g.g\right)\Rightarrow AB^2=BD.BC\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow\frac{AC^2}{AB^2}=\frac{DC.BC}{BD.BC}=\frac{DC}{BD}\Rightarrow\frac{AC^4}{AB^4}=\frac{DC^2}{BD^2}\left(5\right)\)
C/m: \(\Delta DAC\sim\Delta EDC\left(g.g\right)\Rightarrow DC^2=CE.AC\left(3\right)\)
\(\Delta DBA\sim\Delta FBD\left(g.g\right)\Rightarrow BD^2=BF.AB\left(4\right)\)
\(\left(3\right)\left(4\right)\Rightarrow\frac{DC^2}{BD^2}=\frac{CE.AC}{BF.AB}\left(6\right)\)
\(\left(5\right)\left(6\right)\Rightarrow\frac{AC^4}{AB^4}=\frac{CE.AC}{BF.AB}\Rightarrow\frac{AC^3}{AB^3}=\frac{CE}{BF}\Rightarrowđpcm\)