let $a,b,c$ be real numbers (not necessarily positive) for which $\min(a+b,b+c,c+a)\ge 2$. Show that $(a+b+c)^3 \ge a^3 + b^3 + c^3 + 24$

Question

Can anyone solve the following problem:

Let $a,b,c$ be real numbers (not necessarily positive) for which $\min(a+b,b+c,c+a) \ge 2$.

Show that $(a+b+c)^3 \ge a^3 + b^3 + c^3 + 24$

For anybody that doesn't know, the "min" part means that $a+b$, $b+c$ and $c+a$ are all equal to or greater than $2$.

Answer 1

Because $$(a+b+c)^3-a^3-b^3-c^3=3(a+b)(a+c)(b+c)\geq24.$$