\(a+b+c+ab+bc+ca=6abc\)
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)
Đặt \(\left\{{}\begin{matrix}\frac{1}{a}=x\\\frac{1}{b}=y\\\frac{1}{c}=z\end{matrix}\right.\)\(\Rightarrow x+y+z+xy+yz+zx=6\)
CM \(P=x^2+y^2+z^2\ge3\)
\(x^2+1\ge2x;y^2+1\ge2y;z^2+1\ge2z\)
\(2x^2+2y^2+2z^2\ge2xy+2yz+2zx\)
Cộng vế với vế
\(\Rightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+zx\right)=12\)
\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge9\)
\(\Rightarrow x^2+y^2+z^2\ge3\left(đpcm\right)\)
Vậy dấu "=" xảy ra khi \(x=y=z=1\) hoặc \(a=b=c=1\)
\(a+b+c+ab+bc+ca=6abc\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6\)
Ta lại có:
\(\frac{1}{a^2}+1+\frac{1}{b^2}+1+\frac{1}{c^2}+1-3\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}-3\)
\(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}\ge\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}\)
Cộng vế với vế:
\(\frac{3}{a^2}+\frac{3}{b^2}+\frac{3}{c^2}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)-3\)
\(\Leftrightarrow3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge6.2-3=9\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)
Dấu "=" xảy ra khi \(a=b=c=1\)