Đặt \(\left(a;2b;3c\right)=\left(x;y;z\right)\Rightarrow x+y+z=3\)
\(Q=\dfrac{x+1}{1+y^2}+\dfrac{y+1}{1+z^2}+\dfrac{z+1}{1+x^2}\)
Ta có:
\(\dfrac{x+1}{1+y^2}=x+1-\dfrac{\left(x+1\right)y^2}{1+y^2}\ge x+1-\dfrac{\left(x+1\right)y^2}{2y}=x+1-\dfrac{\left(x+1\right)y}{2}\)
Tương tự:
\(\dfrac{y+1}{1+z^2}\ge y+1-\dfrac{\left(y+1\right)z}{2}\) ; \(\dfrac{z+1}{1+x^2}\ge z+1-\dfrac{\left(z+1\right)x}{2}\)
Cộng vế:
\(Q\ge\dfrac{x+y+z}{2}+3-\dfrac{1}{2}\left(xy+yz+zx\right)\)
\(Q\ge\dfrac{x+y+z}{2}+3-\dfrac{1}{6}\left(x+y+z\right)^2=\dfrac{3}{2}+3-\dfrac{9}{6}=3\)
\(Q_{min}=3\) khi \(x=y=z=1\) hay \(\left(a;b;c\right)=\left(1;\dfrac{1}{2};\dfrac{1}{3}\right)\)
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