An Inequality in Determinants

Leo Giugiuc posted the following elegant problem at the CutTheKnotMath facebook page. The problem is due to Dan Sitaru, Solution 2 to Leo Giugiuc.

Assume $1\lt 2a\lt 2b\lt 2c.$ Prove or disprove that

$\left|\begin{array}{ccc} 1 & 1 & 1\\ a & b & c\\ a^4 & b^4 & c^4\\ \end{array}\right|\ge \left|\begin{array}{ccc} 1 & 1 & 1\\ a & b & c\\ a^3 & b^3 & c^3. \end{array}\right|.$

Solution 1

The problem reduces to evaluating the determinant:

$D=\left|\begin{array}{ccc} 1 & 1 & 1\\ a & b & c\\ a^4-a^3 & b^4-b^3 & c^4-c^3\\ \end{array}\right|.$

By subtracting the first row - suitably multiplied - from the second and the third, we arrive at an equal determinant

$D=\left|\begin{array}{ccc} 1 & 1 & 1\\ 0 & b-a & c-a\\ 0 & (b^4-b^3)-(a^4-a^3) & (c^4-c^3)-(a^4-a^3)\\ \end{array}\right|=\left|\begin{array}{ccc} 1 & 1 & 1\\ 0 & b-a & c-a\\ 0 & (b-a)B & (c-a)C\\ \end{array}\right|,$

where

$B=b^3+b^2a+ba^2+a^3-b^2-ab-a^2,\\ C=c^3+c^2a+ca^2+a^3-c^2-ac-a^2.$

Now, subtracting the second row multiplied by $B$ from the third one we again get an equal determinant:

$D=\left|\begin{array}{ccc} 1 & 1 & 1\\ 0 & b-a & c-a\\ 0 & 0 & (c-a)(C-B)\\ \end{array}\right|=(b-a)(c-a)(C-B).$

We find that $C-B=(c-b)X,$ where

$\displaystyle\begin{align} X&=c^2+cb+b^2+ca+ba+a^2-c-b-a\\ &=\left(a-\frac{1}{2}\right)^{2}+\left(b-\frac{1}{2}\right)^{2}+\left(c-\frac{1}{2}\right)^{2}+ab+bc+ca-\frac{3}{4}. \end{align}$

The squares are non-negative which may suggest that the same is true of the whole expression. This is indeed so because, by the AM-GM inequality,

$\displaystyle\frac{1}{3}(ab+bc+ca)\ge\sqrt[3]{a^2b^2c^2}\ge\sqrt[3]{2^{-6}}=\frac{1}{4}.$

The equality could be attained only when $\displaystyle a=b=c=\frac{1}{2}$ which never happens due to $a\lt b\lt c.$ It follows that $D$ equals $(b-a)(c-a)(c-b)$ times a positive expression and is, thus, itself positive.

Solution 2

Let $\displaystyle f:\left(\frac{1}{2},\infty\right)\leftarrow\mathbb{R}$ defined by $f(x)=x^4-x^3.$ It is easily seen that $f''(x)=6x(2x-1)\gt 0,$ for $\displaystyle x\gt\frac{1}{2}.$ On the other hand, the difference of the two determinants, $\Delta_1$ and and $\Delta_2$ can be written as

$\Delta_1-\Delta_2=\left|\begin{array}{ccc} 1 & 1 & 1\\ a & b & c\\ f(a) & f(b) & f(c)\\ \end{array}\right|=\left|\begin{array}{ccc} 1 & 0 & 0\\ a & b-a & c-b\\ f(a) & f(b)-f(a) & f(c)-f(b) \end{array}\right|.$

By the Mean Value Theorem, $f(b)-f(a)=f'(u)(b-a)$ and $f(c)-f(b)=f'(v)(c-b),$ where $a\lt u\lt b\lt v\lt c.$ It follows that

$\Delta_1-\Delta_2==(b-a)(c-b)(f'(v)-f'(u))\gt 0,$

because $f'$ is strictly increasing on $\displaystyle\left(\frac{1}{2},\infty\right).$


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