Problem 1 from the 2017 Canada MO

Problem

Problem 1 from the 2017 Canada MO

Solution 1

Introduce $\displaystyle x=\frac{a}{b-c},$ $\displaystyle y=\frac{b}{c-a},$ $\displaystyle z=\frac{c}{a-b}.$ On an earlier occasion we proved that $xy+yz+zx=-1.$ It follows that

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

To prove that the inequality is strict we shall derive a contradiction from $\displaystyle \sum_{cycl}\left(\frac{a}{b-c}\right)^2=2.$ The consequence of such an assumption would be

$\displaystyle 0=x+y+z=\sum_{cycl}\frac{a}{b-c}=0,$

or,

$a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)=0.$

This is the case of equality in Schur's inequality, so that either $a=b=c$ or one of them $0$ while the other two equal, contradicting the conditions of the problem.

Solution 2

More directly, from Schur's inequality

$a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)\gt 0$

which is strict under the problem's conditions, we derive

$\displaystyle \sum_{cycl}\frac{a}{b-c}\gt 0.$

Squaring and taking into account that $\displaystyle \sum_{cycl}\frac{ab}{(a-c)(c-b)}=-1,$ we obtain the required inequality

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

Solution 3

The left-hand side is symmetric with respect to $a,$ $b,$ $c.$ Hence, we may assume that $a \gt b gt c = 0.$ Note that replacing $(a, b, c)$ with $(a - c, b - c, 0)$ lowers the value of the left-hand side, since the numerators of each of the fractions would decrease and the denominators remain the same. Therefore, to obtain the minimum possible value of the left-hand side, we may assume that $c = 0.$

Then the left-hand side becomes

$\displaystyle \frac{a^2}{b^2}+\frac{b^2}{a^2},$

which yields, by the Arithmetic Mean - Geometric Mean Inequality,

$\displaystyle \frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2\sqrt{\frac{a^2}{b^2}\cdot\frac{b^2}{a^2}}=2,$

with equality if and only if $\displaystyle \frac{a^2}{b^2}=\frac{b^2}{a^2},$ or equivalently, $a^4=b^4.$ Since $a,b \ge 0,$ $a = b.$ But since no two of $a,$ $b,$ $c$ are equal, $a \ne b.$ Hence, equality cannot hold. This yields

$\displaystyle \frac{a^2}{b^2}+\frac{b^2}{a^2}\gt 2.$

Ultimately, this implies the desired inequality.

Solution 4

WLOG, let $a\gt b\gt c$. Let $a-b=x$ and $b-c=y$. Thus,

$\displaystyle \begin{align} \left(\frac{a}{b-c}\right)^2+\left(\frac{b}{c-a}\right)^2+\left(\frac{c}{a-b}\right)^2& =\left(\frac{c+x+y}{y}\right)^2+\left(\frac{c+y}{x+y}\right)^2+\left(\frac{c}{x}\right)^2 \\ &\geq \left(\frac{x+y}{y}\right)^2+\left(\frac{y}{x+y}\right)^2\\ &\gt 2, \end{align}$

by the AM-GM inequality. Note, $x+y>y$. Thus, the last inequality is a strict one.

Acknowledgment

Leo Giugiuc has kindly communicated to me this problem, along with his two solutions. This is problem 1 from the 2017 Canada Mathematical Olympiad. The complete list of the olympiad problems has been posted at the Crux Mathematicorum facebook page. Solution 3 is the official solution; Solution 4 is by Amit Itagi.

 

Cyclic inequalities in three variables

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