Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=1\)
\(VT=\sqrt{\frac{yz}{x+yz}}+\sqrt{\frac{zx}{y+zx}}+\sqrt{\frac{xy}{z+xy}}\)
Ta có: \(\sqrt{\frac{yz}{x+yz}}=\sqrt{\frac{yz}{x\left(x+y+z\right)+yz}}=\sqrt{\frac{yz}{\left(x+y\right)\left(x+z\right)}}\le\frac{1}{2}\left(\frac{y}{x+y}+\frac{z}{x+z}\right)\)
Tương tự: \(\sqrt{\frac{zx}{y+zx}}\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{z}{y+z}\right)\) ; \(\sqrt{\frac{xy}{z+xy}}\le\frac{1}{2}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)\)
Cộng vế với vế:
\(\Rightarrow VT\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{y}{x+y}+\frac{y}{y+z}+\frac{z}{y+z}+\frac{x}{z+x}+\frac{z}{z+x}\right)=\frac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\) hay \(a=b=c=3\)
