Dat \(\left(a,b,c\right)=\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\left(a,b,c>0,abc=1\right)\)
Ta co \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Rightarrow\frac{3}{ab+bc+ca}\ge\frac{9}{\left(a+b+c\right)^2}\left(1\right)\)
BDT phu \(1+\frac{3}{ab+bc+ca}\ge\frac{6}{a+b+c}\left(2\right)\)
Do (1) nen (2) tuong duong voi
\(1+\frac{9}{\left(a+b+c\right)^2}\ge\frac{6}{a+b+c}\Leftrightarrow\left(1-\frac{3}{a+b+c}\right)^2\ge0\left(dung\right)\)
Suy ra (2) duoc chung minh
Do \(abc=1\Rightarrow\hept{\begin{cases}ab=\frac{1}{xy}=\frac{xyz}{xy}=z\\bc=x\\ca=y\end{cases}}\)
nen (2) tuong duong \(1+\frac{3}{x+y+z}\ge\frac{6}{xy+yz+zx}\)
=> \(\frac{1}{x+y+z}\ge\frac{1}{3}\left(\frac{6}{x+y+z}-1\right)=\frac{2}{x+y+z}-\frac{1}{3}\)
Suy ra \(P\ge\frac{2}{x+y+z}-\frac{1}{3}-\frac{2}{x+y+z}=-\frac{1}{3}\)
Dau = xay ra khi x=y=z=1
