Áp dụng bất đẳng thức Cauchy, ta có:

\(\begin{align*}
♥ \dfrac{bc}{a}+\dfrac{ca}{b}&\geq2\sqrt{\dfrac{bc}{a}\cdot\dfrac{ca}{b}}\\
\Leftrightarrow\dfrac{bc}{a}+\dfrac{ca}{b}&\geq2\sqrt{c^2}\\
\Leftrightarrow\dfrac{bc}{a}+\dfrac{ca}{b}&\geq2c\tag1\\
♥ \dfrac{bc}{a}+\dfrac{ab}{c}&\geq2\sqrt{\dfrac{bc}{a}\cdot\dfrac{ab}{c}}\\
\Leftrightarrow\dfrac{ab}{c}+\dfrac{ca}{b}&\geq2\sqrt{b^2}\\
\Leftrightarrow\dfrac{ab}{c}+\dfrac{ca}{b}&\geq2b\tag2\\
♥ \dfrac{ca}{b}+\dfrac{ab}{c}&\geq2\sqrt{\dfrac{ca}{b}\cdot\dfrac{ab}{c}}\\
\Leftrightarrow\dfrac{ca}{b}+\dfrac{ab}{c}&\geq2\sqrt{a^2}\\
\Leftrightarrow\dfrac{ca}{b}+\dfrac{ab}{c}&\geq2a\tag3
\end{align*}\)

Cộng (1), (2) và (3) ta được $$\begin{align*}
2\left(\dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}\right)&\geq2(a+b+c)\\
\Leftrightarrow\left(\dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}\right)&\geq(a+b+c)
\end{align*}$$