For $a\geq2$, $b\geq2$ and $c\geq2$, prove that $\left(a^3+b\right)\left(b^3+c\right)\left(c^3+a\right)\geq125 abc$

For $a\geq2$, $b\geq2$ and $c\geq2$, prove that

$$(a^3+b)(b^3+c)(c^3+a)\geq 125 abc.$$

My try:

First I wrote the inequality as

$$\left(a^2+\frac{b}{a}\right) \left(b^2+\frac{c}{b}\right) \left(c^2+\frac{a}{c}\right) \geq 125. $$

Then I noted that

$$a^2+\frac{b}{a}\geq a^2+\frac{2}{a}, \\ b^2+\frac{c}{b}\geq b^2+\frac{2}{b}, \\ c^2+\frac{a}{c}\geq c^2+\frac{2}{c}. $$

But I don't know how this can help me.

$$a^3+b = {a^3\over 4}+{a^3\over 4}+{a^3\over 4}+{a^3\over 4}+b\geq 5\sqrt[5]{a^{12}b\over 2^8}$$

$$b^3+c = {b^3\over 4}+{b^3\over 4}+{b^3\over 4}+{b^3\over 4}+c\geq 5\sqrt[5]{b^{12}c\over 2^8}$$

$$c^3+a = {c^3\over 4}+{c^3\over 4}+{c^3\over 4}+{c^3\over 4}+a\geq 5\sqrt[5]{c^{12}a\over 2^8}$$

So $$\left(a^3+b\right)\left(b^3+c\right)\left(c^3+a\right) \geq 125abc\sqrt[5]{a^8b^8c^8\over 2^{24}}\geq 125abc$$

You did 90% of the work. You can finish it by using a little bit of Calculus.

Let

$$f(x)=x^2+\frac{2}{x}.$$

Since the derivate of $f(x)$ is positive, the function $f(x)$ is increasing for $x\geq2$, and then

$$f(x)\geq f(2)=5.$$

Thereby,

$$a^2+\frac{b}{a}\geq 5, \\ b^2+\frac{c}{b}\geq 5,\\ c^2+\frac{a}{c}\geq 5,$$

which implies that

$$ \left(a^2+\frac{b}{a}\right) \left(b^2+\frac{c}{b}\right) \left(c^2+\frac{a}{c}\right) \geq 125.$$

Another way.

By AM-GM for all $x\geq2$ we obtain: $$x^3+16=x^3+8+8\geq3\sqrt[3]{x^3\cdot8^2}=12x.$$ Thus, $$\prod_{cyc}(a^3+b)\geq\prod_{cyc}(12a-16+b)=\prod_{cyc}(5a+7a+b-16)\geq\prod_{cyc}5a=125abc.$$