<s>[INST] Proofing an inequality Assume we have three points $ a , b , c $ where $a < b < c$ and $b-a=c-b=h $ 
How can I proof for every $x \in [a,b]$ we have $|(x-a)(x-b)(x-c)| \leq \frac{2\sqrt{3}}{9}h^{3}$
 [/INST] $$f(x)=|(x-a)(x-b)(x-c)|=(x-a)(x-b)(x-c)=(x-a)(b-x)(c-x)\tag{1}$$
Introduce substitution: 
$$b-x=u, \quad u\in[0,h]\tag{2}$$
$$x-a=(b-a)-(b-x)=h-u\tag{3}$$
$$c-x=(c-b)+(b-x)=h+u\tag{4}$$
Now replace (2,3,4) into (1):
$$f(u)=(h-u)u(h+u)=u(h^2-u^2)=uh^2-u^3\tag{5}$$
$$f'(u)=h^2-3u^2=0\implies u=\frac{h\sqrt{3}}{3}\tag{6}$$
$$f(u)|_{u=\frac{h\sqrt{3}}{3}}=\frac{2\sqrt{3}}{9}h^{3}\tag{7}$$
Obviously $f''(u)=-3u^2<0$ for all values of $u\ne0$. This proves that (7) is indeed the maximum value.
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