Linear Algebra Tools for Proving Inequalities

The following is a summary of two presentations Leo Giugiuc made in March 2015 at two local conferences (Romania). The essence of the talks was to present elements of linear algebra as a powerful tool for proving inequalities. The solutions to the two problems described below make use of the well-known result from linear algebra:


A matrix equation $Ax=0,\;$ with a $n\times n\;$ matrix $A\;$ has a non-trivial solution iff the determinant $|A|=0.$

We apply the lemma to solve

Problem 1 (Vasile Cirtoaje)

[a/(b-c)]^2+[b/(c-a)]^2+[c/(a-b)]^2 is not less than 2


Consider matrix $\displaystyle A=\left(\begin{array}{rrr}1 & -x & x\\y & 1 & -y\\-z & z & 1\end{array}\right),$ where $\displaystyle x=\frac{a}{b-c},\;$ $\displaystyle y=\frac{b}{c-a},\;$ $\displaystyle z=\frac{c}{a-b}.\;$ By direct verification,

$\displaystyle \left(\begin{array}{rrr}1 & -x & x\\y & 1 & -y\\-z & z & 1\end{array}\right)\left(\begin{array}{c}a\\b\\c\end{array}\right)=\left(\begin{array}{c}0\\0\\0\end{array}\right).$

It follows that $|A|=xy+yz+zx+1=0,\;$ or $xy+yz+zx=-1.$ We have,

$\displaystyle\begin{align} f &= \left(\frac{a}{b-c}\right)^2+\left(\frac{b}{c-a}\right)^2+\left(\frac{c}{a-b}\right)^2- 2\\ &=x^2+y^2+z^2-2\\ &=x^2+y^2+z^2+2(xy+yz+zx)\\ &=(x+y+z)^2\ge 0. \end{align}$

Now, let's reverse the definition and choose $x,y,z\;$ that satisfy $x+y+z=0\;$ and $xy+yz+zx=-1\;$ but, otherwise, arbitrary. The above system then will have a non-trivial solution $\left(\begin{array}{c}a\\b\\c\end{array}\right)\;$ for which the inequality at hand becomes an equality.

Problem 2

Let $a,b,c\;$ be pairwise distinct real numbers. Then

$\displaystyle\left(\frac{a}{a-b}+3\right)^2+\left(\frac{b}{b-c}+3\right)^2+\left(\frac{c}{c-a}+3\right)^2\ge 25.\;$


Consider matrix $\displaystyle A=\left(\begin{array}{rrr}1-x & x & 0\\0 & 1-y & y\\z & 0 & 1-z\end{array}\right),$ where $\displaystyle x=\frac{a}{a-b},\;$ $\displaystyle y=\frac{b}{b-c},\;$ $\displaystyle z=\frac{c}{c-a}.\;$ By direct verification,

$\displaystyle \left(\begin{array}{rrr}1-x & x & 0\\0 & 1-y & y\\z & 0 & 1-z\end{array}\right)\left(\begin{array}{c}a\\b\\c\end{array}\right)=\left(\begin{array}{c}0\\0\\0\end{array}\right).$

It follows that $|A|=xy + yz + zx - (x + y + z) + 1=0,\;$ or $xy+yz+zx=(x + y + z) - 1.$ Define $s=x+y+z.\;$ Then

$\displaystyle\begin{align} f &= \left(\frac{a}{a-b}+3\right)^2+\left(\frac{b}{b-c}+3\right)^2+\left(\frac{c}{c-a}+3\right)^2-25\\ &=(x+3)^2+(y+3)^2+(z+3)^2-25\\ &=x^2+y^2+z^2+6s+2\\ &=s^2-2(xy+yz+zx)+6s+2\\ &=s^2 - 2(s - 1) + 6s + 2\\ &=s^2+4s+4\\ &=(s+2)^2\ge 0. \end{align}$

As in Problem 1, the equality is achieved for infinitely many triples $a,b,c.$

Below a few inequalities valid under the same conditions as in the above two problems:

  1. $\displaystyle\left(\frac{a}{a-b}\right)^2+\left(\frac{b}{b-c}\right)^2+\left(\frac{c}{c-a}\right)^2\ge 1;$

  2. $\displaystyle\left(\frac{a}{a-b}+k\right)^2+\left(\frac{b}{b-c}+k\right)^2+\left(\frac{c}{c-a}+k\right)^2\ge 2k^2+2k+1;$

  3. $\displaystyle\left(\frac{a+b}{a-b}\right)^2+\left(\frac{b+c}{b-c}\right)^2+\left(\frac{c+a}{c-a}\right)^2\ge 2;$

  4. $\displaystyle\left|\frac{a+b}{a-b}\right|+\left|\frac{b+c}{b-c}\right|+\left|\frac{c+a}{c-a}\right|\ge 2\;$ (Michael Rozenberg, Israel.)

As a side note, concerning the first problem, Inder Jeet Taneja, came up with the factoring:

$\displaystyle\begin{align}X&=\left(\frac{a}{b-c}\right)^2+\left(\frac{b}{c-a}\right)^2+\left(\frac{c}{a-b}\right)^2- 2\\ &=\frac{(a^3-a^2b-a^2c-ab^2+3abc-ac^2+b^3-b^2c-bc^2+c^3)^2}{(b-c)^2(c-a)^2(a-b)^2} \end{align}$


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