\(a+b+c=abc\Leftrightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)
Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\Rightarrow xy+yz+zx=1\)
\(VT=\dfrac{x^3}{z}+\dfrac{y^3}{x}+\dfrac{z^3}{y}=\dfrac{x^4}{xz}+\dfrac{y^4}{xy}+\dfrac{z^4}{yz}\)
\(VT\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{xy+yz+zx}\ge\dfrac{\left(xy+yz+zx\right)^2}{xy+yz+zx}=1\)
Dấu "=" xảy ra khi \(a=b=c=...\)
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