\(\dfrac{bc}{a\left(1+bc\right)}=\dfrac{bc}{a+abc}=\dfrac{bc}{a+b+a+c}\le\dfrac{1}{4}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right)\)
Tương tự và cộng lại:
\(P\le\dfrac{1}{4}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}+\dfrac{ac}{a+b}+\dfrac{ac}{b+c}+\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)=\dfrac{a+b+c}{4}\) (đpcm)
Chứng minh BĐT vế trái:
Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\) \(\Rightarrow\left\{{}\begin{matrix}x+y+z\ge\sqrt{3\left(xy+yz+zx\right)}=\sqrt{3}\\1\ge3\sqrt[3]{\left(xyz\right)^2}\Rightarrow3xyz\le\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)
\(P=\dfrac{x}{1+yz}+\dfrac{y}{1+zx}+\dfrac{z}{1+xy}=\dfrac{x^2}{x+xyz}+\dfrac{y^2}{y+xyz}+\dfrac{z^2}{z+xyz}\)
\(P\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+3xyz}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+\dfrac{1}{\sqrt{3}}}\)
Nên ta chỉ cần chứng minh: \(\dfrac{t^2}{t+\dfrac{1}{\sqrt{3}}}\ge\dfrac{3\sqrt{3}}{4}\) (với \(t=x+y+z\ge\sqrt{3}\))
\(\Leftrightarrow t^2\ge\dfrac{3\sqrt{3}}{4}t+\dfrac{3}{4}\) \(\Leftrightarrow\left(t-\sqrt{3}\right)\left(4t+\sqrt{3}\right)\ge0\) (luôn đúng)
