Problem From the 2016 IMO Shortlist

Problem

Problem From the 2016 IMO Shortlist

Solution 1

The required inequality is equivalent to

$\displaystyle \ln(a^2+1)+\ln(b^2+1)+\ln(c^2+1)\le 3\ln\left[\left(\frac{a+b+c}{3}\right)^2+1\right].$

Consider the function $f:\,(0,\infty)\to\mathbb{R},\,$ defined by $f(x)=\ln(x^2+1).\,$ We have $\displaystyle f'(x)=\frac{2x}{x^2+1}\,$ and $\displaystyle f''(x)=-\frac{2(x^2-1)}{(x^2+1)^2}.\,$ It follows that function $f\,$ is concave on $\displaystyle \left[\frac{1}{2},\infty\right).$

Let, WLOG, $a\le b\le c.\,$ Then it could not be that $b\lt 1,\,$ for, otherwise, $a\lt 1\,$ also and then $ab\lt 1,\,$ in contradiction with the constraint. Thus, $b\ge 1.\,$ Since, by the assumption, $c\ge b,\,$ c\ge 1.\,$ We consider two cases:

  1. $\mathbf{a\ge 1}$

    By Jensen's inequality, $\displaystyle f(a)+f(b)+f(c)\le 3f\left(\frac{a+b+c}{3}\right),\,$ i.e.,

    $\displaystyle \ln(a^2+1)+\ln(b^2+1)+\ln(c^2+1)\le 3\ln\left[\left(\frac{a+b+c}{3}\right)^2+1\right],$

    as required.

  2. $\mathbf{0\lt a\le 1}$

    By Jensen's inequality, $\displaystyle f(b)+f(c)\le 2f\left(\frac{b+c}{2}\right),\,$ so that,

    $\displaystyle\small{\ln(a^2+1)+\ln(b^2+1)+\ln(c^2+1)\le \ln(a^2+1)+2\ln\left[\left(\frac{b+c}{2}\right)^2+1\right]}.$

    Suffice it to prove that

    $\displaystyle \ln(a^2+1)+2\ln\left[\left(\frac{b+c}{2}\right)^2+1\right]\le3\ln\left[\left(\frac{a+b+c}{3}\right)^2+1\right].$

    Since $ab,ac\ge 1,\,$ $\displaystyle a\left(\frac{b+c}{2}\right)\ge 1.\,$ Let $\displaystyle \frac{b+c}{2}=t.\,$ Then $\displaystyle t\ge \frac{1}{a}.\,$ We'll show that, for all $\displaystyle s\ge\frac{1}{a},\,$

    $\displaystyle \ln(a^2+1)+2\ln(s^2+1)\le 3\ln\left[\left(\frac{a+2s}{3}\right)^2+1\right]$

    Consider the function $\displaystyle g:\,\left[\frac{1}{a},\infty\right)\to\mathbb{R},\,$ defined by

    $\displaystyle g(x)=3\ln\left[\left(\frac{a+2s}{3}\right)^2+1\right]-2\ln(s^2+1)-\ln(a^2+1).$

    Then

    $\displaystyle\begin{align} g'(x)&=\frac{\displaystyle 4\left(\frac{a+2x}{3}\right)}{\displaystyle \left(\frac{a+2x}{3}\right)^2+1}-\frac{4x}{x^2+1}\\ &=2\left(f'\left(\frac{a+2x}{3}\right)-f'(x)\right). \end{align}$

    But $a+x\ge 2\sqrt{ax}\ge 2\,$ and $\displaystyle x\ge\frac{1}{a}\ge 1,\,$ implying $\displaystyle \frac{a+2x}{3}\ge 1.\,$ Also, $\displaystyle \frac{a+2x}{3}\le x.\,$ Now, since $f'\,$ is decreasing on $[a,\infty),\,$ $\displaystyle f'\left(\frac{a+2x}{3}\right)\ge f'(x),\,$ implying $g'(x)\ge 0,\,$ for all $\displaystyle x\ge\frac{1}{a},\,$ so that

    $\displaystyle\begin{align}\min g&=g\left(\frac{1}{a}\right)\\ &=3\ln(a^4+13a^2+4)-3\ln(a^2+1)-\ln a^2-6\ln 3. \end{align}$

    Finally, consider the function $h:\,(0,1]\to\mathbb{R},\,$ defined by

    $h(x)=3\ln(x^2+13x+4)-3\ln(x+1)-\ln(x)-6\ln(3).$

    Then $\displaystyle h'(x)=\frac{(x-1)^2(x-2)}{x(x+1)(x^2+13x+4)}\le 0,\,$ for all $x\in (0,1],\,$ implying $h(a^2)\ge h(1)=0,\,$ and this completes the proof.

Solution 2

Note that, if $\sqrt{xy}\ge 1\,$ then also $\displaystyle \frac{x+y}{2}\ge\sqrt{xy}\ge 1.\,$ Thus,

$\displaystyle \frac{a+b}{2}\ge 1,\,\frac{b+c}{2}\ge 1,\,\frac{c+a}{2}\ge 1,$

implying, in particular, $\displaystyle a+b+c=\frac{a+b}{2}+\frac{b+c}{2}+\frac{c+a}{2}\ge 3.$

The inequality is equivalent to

(1)

$9\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)}\le (a+b+c)^2+9.$

Now we prove

Lemma

Prove that, for positive real numbers $x,y,\,$ subject to $xy\ge 1,$

$\displaystyle (x^2+1)(y^2+1)\le\left(1+\left(\frac{x+y}{2}\right)^2\right)^2.$

Indeed, $(x^2+1)(y^2+1)=1+x^2+y^2+x^2y^2=1+s^2-2p+p^2,\,$ where $s=x+y\,$ and $p=xy\ge 1.$

Define function $\displaystyle f:\,\left[1,\frac{s^2}{4}\right]\to\mathbb{R}\,$ by $f(p)=1+s^2-2p+p^2.\,$ $f'(p)=2p-2\ge 0,\,$ making $f\,$ increasing, so that

$\displaystyle f(p)\le f\left(\frac{s^2}{4}\right)=1-s^2-2\frac{s^2}{4}+\frac{s^4}{16}=\left(1+\frac{s^2}{4}\right)^2,$

which is exactly the lemma's claim.

Because of the constraint, we may assume that at least one of $a,b,c\,$ is not less than $1,\,$ say, $c\ge 1.\,$ Then also $\displaystyle c\cdot\frac{a+b+c}{3}\ge 1,\,$ and the lemma shows that

$\displaystyle (1+c^2)\left(\left(\frac{a+b+c}{3}\right)^2+1\right)\le\left(1+\left(\frac{\displaystyle c+\frac{a+b+c}{3}}{2}\right)^2\right)^2.$

Also

$\displaystyle \begin{align} &\left(1+\left(\frac{a+b}{2}\right)^2\right)^2\cdot\left(1+\left(\frac{\displaystyle c+\frac{a+b+c}{3}}{2}\right)^2\right)^2\\ &\qquad=\left[\left(1+\left(\frac{a+b}{2}\right)^2\right)\cdot\left(1+\left(\frac{\displaystyle c+\frac{a+b+c}{3}}{2}\right)^2\right)\right]^2\\ &\qquad\le\left[1+\left(\frac{\displaystyle \frac{a+b}{2}+\frac{\displaystyle c+\frac{a+b+c}{3}}{2}}{2}\right)^2\right]^2\\ &\qquad=\left[1+\left(\frac{a+b+c}{3}\right)^2\right]^4. \end{align}$

Thus we see that

$\displaystyle (a^2+1)(b^2+1)(c^2+1)\left(1+\left(\frac{a+b+c}{3}\right)^2\right)\le\left[1+\left(\frac{a+b+c}{3}\right)^2\right]^4,$

or

$\displaystyle (a^2+1)(b^2+1)(c^2+1)\le\left[1+\left(\frac{a+b+c}{3}\right)^2\right]^3.$

Acknowledgment

My thanks go to Leo Giugiuc who has kindly communicated to me this problem from the 2016 IMO Shortlist, along with a solution of his. Solution 2 is by Marian Dinca who also observed that the problem admits a generalization:

Prove that, for positive real numbers $a_k,\,$ $k=1,\ldots,n,\,$ subject to $\min\{a_ka_{k+1}\}\ge 1,\,$ where $a_{n+1}=a_1,$

$\displaystyle \sqrt[n]{\prod_{k=1}^n(a_k^2+1)}\le\left(\frac{\displaystyle \sum_{k=1}^na_k}{\displaystyle n}\right)^2+1.$

 

|Contact| |Up| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

72392305