Apply inequality cauchy, we have:

\(\left\{{}\begin{matrix}\left(b+c-a\right)+\left(a+c-b\right)\ge2\sqrt{\left(b+c-a\right)\left(a+c-b\right)}\\\left(a+c-b\right)+\left(a+b-c\right)\ge2\sqrt{\left(a+c-b\right)\left(a+b-c\right)}\\\left(a+b-c\right)+\left(b+c-a\right)\ge2\sqrt{\left(a+b-c\right)\left(b+c-a\right)}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2c\ge2\sqrt{\left(b+c-a\right)\left(a+c-b\right)}\\2a\ge2\sqrt{\left(a+c-b\right)\left(a+b-c\right)}\\2b\ge2\sqrt{\left(a+b-c\right)\left(b+c-a\right)}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}c^2\ge\left(b+c-a\right)\left(a+c-b\right)\\a^2\ge\left(a+c-b\right)\left(a+b-c\right)\\b^2\ge\left(a+b-c\right)\left(b+c-a\right)\end{matrix}\right.\)

\(\Rightarrow\left(abc\right)^2\ge\left[\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)\right]^2\)

\(\Rightarrow abc\ge\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)\)

Apply inequality cauchy, we have:

⎧⎩⎨⎪⎪⎪⎪(b+c−a)+(a+c−b)≥2(b+c−a)(a+c−b)−−−−−−−−−−−−−−−−−√(a+c−b)+(a+b−c)≥2(a+c−b)(a+b−c)−−−−−−−−−−−−−−−−−√(a+b−c)+(b+c−a)≥2(a+b−c)(b+c−a)−−−−−−−−−−−−−−−−−√

⇒⎧⎩⎨⎪⎪⎪⎪2c≥2(b+c−a)(a+c−b)−−−−−−−−−−−−−−−−−√2a≥2(a+c−b)(a+b−c)−−−−−−−−−−−−−−−−−√2b≥2(a+b−c)(b+c−a)−−−−−−−−−−−−−−−−−√

⇒⎧⎩⎨⎪⎪c2≥(b+c−a)(a+c−b)a2≥(a+c−b)(a+b−c)b2≥(a+b−c)(b+c−a)

⇒(abc)2≥[(b+c−a)(a+c−b)(a+b−c)]2

⇒abc≥(b+c−a)(a+c−b)(a+b−c)