Điều kiện a;b;c dương
Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=3\)
\(\Rightarrow P=\dfrac{y}{2x^2+1}+\dfrac{z}{2y^2+1}+\dfrac{x}{2z^2+1}\)
Ta có:
\(\dfrac{y}{2x^2+1}=y-\dfrac{2x^2y}{x^2+x^2+1}\ge y-\dfrac{2x^2y}{3\sqrt[3]{x^4}}=y-\dfrac{2}{3}y\sqrt[3]{x^2}\ge y-\dfrac{2}{9}y\left(x+x+1\right)=\dfrac{7}{9}y-\dfrac{4}{9}xy\)
Tương tự và cộng lại...
\(\Rightarrow P\ge\dfrac{7}{9}\left(x+y+z\right)-\dfrac{4}{9}\left(xy+yz+zx\right)\ge\dfrac{7}{9}\left(x+y+z\right)-\dfrac{4}{27}\left(x+y+z\right)^2=1\)
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