# An Inequality with Determinants VI

### Solution

\displaystyle\begin{align} \Delta_1 &= \left|\begin{array}{ccc} \,1 & a & a^3\\ 1 & b & b^3\\ 1 & c & c^3\end{array}\right|^{r_1:=r_1-r_2,\,r_2=r_2-r_3}=\left|\begin{array}{ccc} \,0 & a-b & a^3-b^3\\ 0 & b-c & b^3-c^3\\ 1 & c & c^3\end{array}\right|\\ &=(a-b)(b-c)\left|\begin{array}{cc}\,1 & a^2+ab+b^2\\1 & b^2+bc+c^2\end{array}\right|\\ &=(a-b)(b-c)(c-a)(a+b+c). \end{align}

\displaystyle\begin{align} \Delta_2 &= \left|\begin{array}{ccc} \,a^2 & b^2 & c^2\\ b^2+c^2 & c^2+a^2 & a^2+b^2\\ bc & ca & ab\end{array}\right|^{r_2:=r_1+r_2}\\ &=(a^2+b^2+c^2)\left|\begin{array}{ccc} \,a^2 & b^2 & c^2\\ 1 & 1 & 1\\ bc & ca & ab\end{array}\right|\\ &=\frac{a^2+b^2+c^2}{abc}\left|\begin{array}{ccc} \,a^3 & b^3 & c^3\\a & b & c\\abc&abc&abc\end{array}\right|\\ &=(a^2+b^2+c^2)\left|\begin{array}\,a^3&b^3&c^3\\a&b&c&\\1&1&1\end{array}\right|\\ &=(a^2+b^2+c^2)\Delta_1. \end{align}

Thus, we have

\displaystyle\begin{align} \frac{\Delta_1-\Delta_2}{(a-b)(b-c)(c-a)}&=\small{(a+b+c)+(a+b+c)(a^2+b^2+c^2)}\\ &\ge 3(abc)^{\frac{1}{3}}+3(abc)^{\frac{1}{3}}(a^2+b^2+c^2)\\ &\ge 3(abc)^{\frac{1}{3}}\cdot 4(a^2b^2c^2)^{\frac{1}{4}}\\ &= 12(abc)^{\frac{5}{6}}. \end{align}

### Acknowledgment

The inequality from his book "Ice Math" (Problem 026) has been kindly shared at the CutTheKnotMath facebook page by Dan Sitaru, along with a solution by Ravi Prakash.