Find the largest number \(k>0\) with the property that for any \(a, b, c\ge 0\) the following inequality holds: \[(a-b)(b-c)(c-a)\le \frac{(a+b+c)^3}{k}.\]
Tags: inequality calculus
We'll prove that the wanted maximum value \(k_\max\) is \(6\sqrt 3\).
It suffices to consider the case \((a-b)(b-c)(c-a)\ge 0\). WLOG \(a\le b\le c\). Then exist \(x, y\ge 0\) such that \(b=a+x, c=a+x+y\). If some two of \(a, b, c\) are equal then the LHS of the inequality is \(0\) so we may discuss only the case \(a< b< c\). We need to find \[k_\max=\min_{\substack{x, y\>0\\ a\ge 0}}\frac{(3a+2x+y)^3}{xy(x+y)}.\] Note that putting \(a=0\) makes the expression in \(\min\) not larger so we may search the minimum of \[\frac{(2x+y)^3}{xy(x+y)}.\]
We'll try to find such a \(t\in(0, 1)\) that when applying the Cauchy inequality in the form \begin{equation} \label{eq} \frac{(2x+y)^3}{27}=\frac{((2-t)x+(1-t)y+t(x+y))^3}{27}\ge (2-t)(1-t)t\cdot xy(x+y)\tag{*} \end{equation} the value of \(f(t)=(2-t)(1-t)t\) is the maximal. Differentiating \(f\) \[f'(t)=3\left(t-1-\frac{1}{\sqrt 3}\right)\left(t-1+\frac{1}{\sqrt 3}\right)\] we see that on \((0, 1)\) \(f\) attains its maximum at \(t_\max=1-\frac{1}{\sqrt 3}\). Inserting this value we find \[k_\max\ge 27f(t_\max)=6\sqrt 3.\] \eqref{eq} becomes equality iff \[(2-t)x=(1-t)y=t(x+y)\] which is equivalent to \(x:y=1:1+\sqrt 3\) which in turn is equivalent to \(b:c=1:2+\sqrt 3\) in case of \(a=0\). Inserting \((a, b, c)=(0, 1, 2+\sqrt 3)\) into the inequality we make sure that \(k\le 6\sqrt 3\) finishing the proof.