# Problem 4142 From Crux Mathematicorum

### Solution 1

From Hölder's inequality:

\displaystyle \begin{align} &\Bigr(\sum_{cycl}\frac{a}{ax+b}\Bigr)\Bigr(\sum_{cycl}a(ax+b)\Bigr)\geq (a+b+c)^2\\ &\sum_{cycl} \frac{a}{ax+b}\geq \frac{(a+b+c)^2}{a(ax+b)+b(bx+c)+c(cx+a)}\\ &\sum_{cycl} \int_0^1 \frac{a}{ax+b}dx\geq (a+b+c)^2 \int_0^1 \frac{dx}{x(a^2+b^2+c^2)+ab+bc+ca}\\ &\sum_{cycl} \ln (ax+b)\Bigr|_0^1\geq \frac{(a+b+c)^2}{a^2+b^2+c^2}\ln \Bigr(x(a^2+b^2+c^2)+ab+bc+ca\Bigr)\Bigr|_0^1\\ &\sum_{cycl} \ln \Bigr(\frac{a+b}{b}\Bigr)\geq \frac{(a+b+c)^2}{a^2+b^2+c^2}\ln \frac{a^2+b^2+c^2+ab+bc+ca}{ab+bc+ca}\\ &\ln \prod\Bigr(1+\frac{a}{b}\Bigr)\geq \frac{(a+b+c)^2}{a^2+b^2+c^2}\ln \Bigr(1+\frac{a^2+b^2+c^2}{ab+bc+ca}\Bigr)\\ &\Bigr(1+\frac{a}{b}\Bigr)\Bigr(1+\frac{b}{c}\Bigr)\Bigr(1+\frac{c}{a}\Bigr)\geq \Bigr(1+\frac{a^2+b^2+c^2}{ab+bc+ca}\Bigr)^{\frac{(a+b+c)^2}{a^2+b^2+c^2}} \end{align}

### Solution 2

Assuming, due to the homogeneity of the original inequality, that $a+b+c=1$ and denoting $p=ab+bc+ca$ and $q=abc,$ we obtain

\displaystyle \begin{align} &a^2+b^2+c^2=1-2p,\\ &\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{(a+b)(b+c)(c+a)}{abc}=\frac{p-q}{q},\\ &1+\frac{a^2+b^2+c^2}{ab+bc+ca}=1+\frac{1-2p}{p}=\frac{1-p}{p}. \end{align}

The original inequality then becomes

$\displaystyle \left(\frac{1-p}{p}\right)^{\frac{1}{1-2p}}\le\frac{p-q}{q}.$

Since $\displaystyle 0\lt q\le\frac{p^2}{3},$ we have $\displaystyle \frac{p}{q}\ge\frac{3}{p},$ and suffice it to prove the inequality

$\displaystyle\left(\frac{1-p}{p}\right)^{\frac{1}{1-2p}}\le\frac{3}{p}-1.$

(Now note that $1=(a+b+c)^2=(a^2+b^2+c^2)+2p\ge 3p,$ by a well-known inequality. Thus, $\displaystyle p\le\frac{1}{3}.$)

For $\displaystyle 0\lt p\le\frac{1}{3},$ this is successively equivalent to

\displaystyle \begin{align} &\frac{1-p}{p}\le\left(\frac{3-p}{p}\right)^{1-2p},\\ &\left(\frac{3-p}{p}\right)^{2p}\le\frac{3-p}{1-p},\\ &\left(\frac{3-p}{p}\right)^{2}\le\left(\frac{\displaystyle\frac{3}{p}-1}{\displaystyle \frac{1}{p}-1}\right)^{\frac{1}{p}}.\\ \end{align}

Denoting $\displaystyle t=\frac{1}{p}\in[3,\infty),$ we obtain the following more convenient equivalent form of the latter inequality:

$\displaystyle (3t-1)^2\le \left(\frac{3t-1}{t-1}\right)^t\;\Leftrightarrow\;t\ln\left(\frac{3t-1}{t-1}\right)\ge 2\ln (3t-1).$

Let $h(t)=t[\ln(3t-1)-\ln(t-1)]-2\ln(3t-1).$ Then

\displaystyle \begin{align} h'(t)&=\ln(3t-1)-\ln(t-1)+t\left(\frac{3}{3t-1}-\frac{1}{t-1}\right)-\frac{6}{3t-1}\\ &=\ln(3t-1)-\ln(t-1)-\frac{1}{t-1}-\frac{5}{3t-1} \end{align}

and

\displaystyle \begin{align} h''(t)&=\frac{3}{3t-1}-\frac{1}{t-1}+\frac{1}{(t-1)^2}+\frac{15}{(3t-1)^2}\\ &=\frac{2(9t^2-14t+7)}{(3t-1)^2(t-1)^2}. \end{align}

Since $h''(t)\gt 0$ for $t\ge 3,$ $h'(t)$ increases on $[3,\infty)$ and, therefore,

$\displaystyle h'(t)\ge h'(3)=\ln 8-\ln 2-\frac{1}{2}-\frac{5}{8}=2\ln 2-\frac{9}{8}\gt 0.$

Hence, $h(t)$ increase on $[3,\infty)$ and, therefore,

$h(t)\ge h(3)=3(\ln 8-\ln 2)-2\ln 8=0.$

Thus $\displaystyle t\left(\frac{3t-1}{t-1}\right)\ge 2\ln(3t-1),$ as desired.

### Solution 3

Recollect that, for $x\ge 1,$ $2^x\ge x+1.$ From here, successively, $(1+x)^{\frac{1}{x}}\le 2,$ $(1+x)^{\frac{2}{x}}\le 4,$ and

(1)

$\displaystyle (1+x)^{1+\frac{2}{x}}\le 4(1+x).$

Let $\displaystyle x=\frac{a^2+b^2+c^2}{ab+bc+ca}.$ Then, $x\ge 1.$ Replacing $x$ in (1), we get

$\displaystyle \left(1+\frac{a^2+b^2+c^2}{ab+bc+ca}\right)^{\frac{(a+b+c)^2}{a^2+b^2+c^2}}\le 4\left(1+\frac{a^2+b^2+c^2}{ab+bc+ca}\right).$

Hence, suffice it to prove that

$\displaystyle 4\left(1+\frac{a^2+b^2+c^2}{ab+bc+ca}\right)\le\frac{(a+b)(b+c)(c+a)}{abc}.$

It can be seen that the latter is equivalent to

(2)

$\small{abd[4(a+b+c)^2-3(ab+bc+ca)]\le (a+b+c)(ab+bc+ca)^2}.$

Since both sides in (2) are homogeneous of degree $5,$ assume, WLOG, that $a+b+c=3.$ Let $ab+bc+ca=3(1-t^2),$ where $0\le t\le 1.$ Then (2) is equivalent to

$abc(3+t^2)\le 3(1-t)^2(1+t)^2.$

But from Vo Quec Ba Can's theorem, $abc\le (1-t)^(1+2t).$ Thus we may prove a stronger inequality, viz.,

$(1-t)^(1+2t)(3+t^2)\le 3(1-t)^2(1+t)^2$

which is equivalent to $(1+2t)(3+t^2)\le 3(1+t^2)$ which reduces to $t^3\le t^2,$ which holds since $0\le t\lt 1.$ This completes the proof.

Note that from the above it follows that equality is attained for $a=b=c.$

### Acknowledgment

This is problem 4142 from the Canadian Crux Mathematicorum (Volume 43(5), May 2017.) The problem was proposed by Dan Sitaru who communicate to the problem and his solution (Solution 1) as LaTeX file, for which I am extremely grateful. Solution 2 is by Arkady Alt; Solution 3 is by Leo Giugiuc.

The first two solutions use elements of Calculus.