Cyclic Inequality In Three Variables From Kvant


Cyclic  Inequality In  Three  Variables From Kvant, problem

Solution 1

WLOG, $0\le a\le b\le c\le 1.$ Since $(1-a)(1-b)\ge 0,$

$a+b\le 1+ab\le 1+2ab.$

It follows that $a+b+c\le a+b+1\le 2+2ab.$ And now from $1+ab\le 1+ca\le 1+bc,$

$\displaystyle \frac{a}{bc+1}+\frac{b}{ca+1}+\frac{c}{ab+1}\le \frac{a+b+c}{1+ab}\le 2.$

Solution 2

Assume, as before, that $0\le a\le b\le c\le 1,$ and consider the left-hand side of the inequality we are to prove as the function of $a$ on the interval $0\le a\le b:$

$\displaystyle F(a)=\frac{a}{bc+1}+\frac{b}{ca+1}+\frac{c}{ab+1}.$

The function is convex so that its maximum occurs at one of the two endpoints of $[0,b].$ It follows that either $F(a)\le F(0)=b+c$ or $\displaystyle F(a)\le F(b)=\frac{2b}{bc+1}+\frac{c}{b^2+1}.$

Similar argument applies to any function $y=f(x)$ that is the sum of functions $y=kx$ and $\displaystyle y=\frac{p}{qx+r},$ with positive $k,p,q,r$ for $x\gt 0.$

Under the given constraints, $b+c\le 2,$ implying that the latter expression (as a function of $c)$ on $[b,1]$ does not exceed one of the two quantities

$\displaystyle \frac{2b}{b^2+1}+\frac{b}{b^2+1}=\frac{3b}{b^2+1}\le\frac{3}{2},$


$\displaystyle \frac{2b}{b+1}+\frac{1}{b^2+1}=2-\frac{2b^2-b+1}{(b+1)(b^2+1)}\le 2.$

It is possible to consider the left-hand side of the inequality at hand over the whole of the cube $[0,1]^3.$ That function of three variables attains its maximum at one of the corners of the cube. By inspection,

$\displaystyle \begin{array}{c|c|c|c}\\ a&b&c& LHS\\ \hline 0&0&1&1\\ 0&1&1&2\\ 1&1&1&3/2 \end{array}$

which makes it clear that the maximum of $2$ is attained at $(0,1,1)$ and permutations.

Solution 3

$\displaystyle f(a,b,c)=a/(bc+1)+b/(ac+1)+c/(ab+1).$ Suppose $a\leq b\leq c.$ As a function of $c$, we have $f''(c)\geq 0$, hence keeping $c$ variable,

$\displaystyle \begin{align} f'(c)\geq f'(b)\;&=-ab/(b^2+1)^2-ab/(ab+1)^2+1/(ab+1)\\ &=-ab(b^2+1)^2+1/(ab+1)^2\\ &=\frac{(b^2+1)^2-ab(ab+1)^2}{(b^2+1)^2(ab+1)^2}\geq 0, \end{align}$

hence the maximum is attained at $c=1$. So

$\displaystyle f=a/(b+1)+b/(a+1)+1/(ab+1).$

As function of $b$, $f''(b)\geq 0$, so keeping $b$ variable,

$\displaystyle \begin{align} f'(b)\geq f'(a)\;&=-a/(a+1)^2+1/(a+1)-a/(a^2+1)^2\\ &=\frac{(a-1)(a^3-a-1)}{(a+1)^2(a^2+1)^2}\geq 0 \end{align}$

so the maximum is at $b=1$. Then $f=a/2+2/(a+1)$ which is decreasing, so $a=0$.


This is problem 1228 from the Russian math magazine Kvant (1990, n 6). The inequality is by D. Fomin. Solutions 1,2 are by D. Fomin and N. Vasilyev; Solution 3 is by Bogdan Lataianu.


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