Áp dụng BĐT AM - GM, ta có:
\(\dfrac{a^3+b^3+c^3}{2abc}+\dfrac{a^2+b^2}{c^2+ab}+\dfrac{b^2+c^2}{a^2+bc}+\dfrac{a^2+c^2}{b^2+ac}\)
\(\ge\dfrac{3\sqrt{a^3b^3c^3}}{2abc}+\dfrac{a^2+b^2}{c^2+\dfrac{a^2+b^2}{2}}+\dfrac{b^2+c^2}{a^2+\dfrac{b^2+c^2}{2}}+\dfrac{a^2+c^2}{b^2+\dfrac{a^2+c^2}{2}}\)
\(\ge\dfrac{3abc}{2abc}+\dfrac{2\left(a^2+b^2\right)}{2c^2+a^2+b^2}+\dfrac{2\left(b^2+c^2\right)}{2a^2+b^2+c^2}+\dfrac{2\left(a^2+c^2\right)}{2b^2+a^2+c^2}\)
\(=\dfrac{3}{2}+2\times\left[\dfrac{a^2+b^2}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}+\dfrac{b^2+c^2}{\left(a^2+b^2\right)+\left(a^2+c^2\right)}+\dfrac{c^2+a^2}{\left(b^2+c^2\right)+\left(b^2+a^2\right)}\right]\) (1)
Đặt \(\left\{{}\begin{matrix}a^2+b^2=x\\b^2+c^2=y\\c^2+a^2=z\end{matrix}\right.\), ta có:
\(\left(1\right)\Leftrightarrow\dfrac{3}{2}+2\times\left(\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\right)\)
\(\ge\dfrac{3}{2}+2\times\dfrac{3}{2}\) (Bất_đẳng_thức_Nesbitt)
\(=\dfrac{9}{2}\left(\text{đ}pcm\right)\)
Dấu "=" xảy ra khi a = b = c