# Another Problem from the 2016 Danubius Contest

### Problem ### Solution 1

The inequality is equivalent to

$\displaystyle 12+4\sum_{cycl}a^2+\sum_{cycl}a^2b^2\le 8+4\sum_{cycl}a^2+2\sum_{cycl}a^2b^2+a^2b^2c^2,$

or,

$\displaystyle 4\le a^2b^2+b^2c^2+c^2a^2+a^2b^2c^2.$

let's denote $bc=x,\,$ $ca=y,\,$ and $ab=z.\,$ Then $x,y,z\gt 0,\,$ $x+y+z=3\,$ and $xyz=a^2b^2c^2.\,$ We have to show that

$x^2+y^2+z^2+xyz\ge 4.$

We can homogenize the inequality:

$\displaystyle 4\left(\frac{x+y+z}{3}\right)^3\le (x^2+y^2+z^2)\left(\frac{x+y+z}{3}\right)+xyz,$

reducing it to $5S_3-3s+3xyz,\,$ where $s_3=x^3+y^3+z^3\,$ and $s=\displaystyle \sum_{cycl}xy(x+y).\,$ From Schur's inequality, $S_3-s+3xyz\ge 0\,$ and, from the well-known inequality $u^3+v^3\ge uv(u+v),\,$ $4S_3-2s\ge 0.\,$ Adding the two up gives the required inequality.

### Solution 2

$\displaystyle\sum_{cycl}\frac{1}{a^2+2}\le 1\,$ is equivalent to $\displaystyle\sum_{cycl}\frac{2}{a^2+2}\le 2\,$ and, further, to $\displaystyle\sum_{cycl}\left(1-\frac{a^2}{a^2+2}\right)\le 2\,$ which is $\displaystyle\sum_{cycl}\frac{a^2}{a^2+2}\ge 1.\,$ To prove this we can use Bergstrom's inequality:

\displaystyle\begin{align} \sum_{cycl}\frac{a^2}{a^2+2}&\ge\frac{(a+b+c)^2}{a^2+b^2+c^2+6}\\ &=\frac{(a^2+b^2+c^2)+2(ab+bc+ca)}{a^2+b^2+c^2+6}\\ &=\frac{(a^2+b^2+c^2)+2\cdot 3}{a^2+b^2+c^2+6}=1. \end{align}

### Solution 3

let's denote $bc=x,\,$ $ca=y,\,$ and $ab=z.\,$ Then $x,y,z\gt 0,\,$ $x+y+z=3.\,$ We have to show that

$x^2+y^2+z^2+xyz\ge 4.$

Assuming $x=\min\{x,y,z\},\,$ $x\le 1,\,$ and we have

\displaystyle \begin{align} x^2+y^2+z^2+xyz-4 &= x^2+(y+z)^2+yz(x-2)-4\\ &\ge x^2+(y+z)^2+\frac{1}{4}(y+z)^2(x-2)-4\\ &\ge x^2+\frac{x+2}{4}(y+z)^2-4\\ &=x^2+{x+2}{4}(3-x)^2-4\\ &=\frac{1}{4}(x-1)^2(x+2)\ge 0. \end{align}

Equality occurs for $a=b=c=1.$

### Solution 4

$\displaystyle c=\frac{3-ab}{a+b}.$

\displaystyle\begin{align} RHS-LHS &= \frac{c^2+1}{c^2+2}-\frac{a^2+b^2+4}{(a^2+2)(b^2+2)}\\ &=\frac{(3-ab)^2+(a+b)^2}{(3-ab)^2+2(a+b)^2}-\frac{a^2+b^2+4}{(a^2+2)(b^2+2)}. \end{align}

Denominators being positive, we'll focus on the numerators:

\begin{align} &(2a^2+2b^2+4)((3-ab)^2+(a+b)^2)\\ &\qquad\qquad\qquad\qquad-(a^2+b^2+4)((3-ab)^2+2(a+b)^2)\\ &\qquad\qquad\qquad\qquad+a^2b^2((3-ab)^2+(a+b)^2)\\ &=(a^2+b^2)(3-ab)^2-4(a^2+b^2)\\ &\qquad\qquad\qquad\qquad-8ab+a^2b^2(a^2+b^2-4ab+9+a^2b^2)\\ &=(a^2+b^2)(a^2b^2-6ab+9-4+a^2b^2)\\ &\qquad\qquad\qquad\qquad-ab(a^3b^3-4a^2b^2+9ab-8)\\ &=(a^2+b^2)(2a^2b^2-6ab+4)+ab(a^3b^3-4a^2b^2+9ab-6)\\ &\qquad\qquad\qquad\qquad+(a-b)^2\\ &=(a^2+b^2)(ab-1)(ab-2)+ab(ab-1)(a^2b^2-3ab+6)\\ &\qquad\qquad\qquad\qquad+(a-b)^2\\ &\ge 2\cdot 2ab(ab-1)(ab-2)+ab(ab-1)(a^2b^2-3ab+6)\\ &\qquad\qquad\qquad\qquad+(a-b)^2\\ &=ab(ab-1)(a^2b^2+ab-2)+(a-b)^2\\ &=ab(ab-1)(ab-1)(ab+2)+(a-b)^2\\ &=ab(ab-1)^2(ab+2)+(a-b)^2\\ &\ge 0. \end{align}

### Solution 5

We make the same substitution as in Solutions 1 and 3 to obtain the problem of proving

$xyz+x^2+y^2+z^2-4\ge 0,$

provided $x+y+z=3\,$ and $x,y,z\gt 0.\,$ Let $x=k+m,\,$ $y=k-m,\,$ $z=3-x-y=3-2k.\,$ The positivity of $x,y,\,$ and $z\,$ implies

$\displaystyle \frac{3}{2}\gt k\gt 0\,$ and $k\gt m\gt -k.$

We have

\begin{align} f(k,m) &:= (k+m)(k-m)(3-2k)+(k+m)^2\\ &\qquad\qquad\qquad\qquad+(k-m)^2+(3-2k)^2-4\\ &=[(5-2k)(k-1)^2]+[(2k-1)m^2]. \end{align}

Note: The first term (the first square bracket) is non-negative over the alowed range of $k,\,$ becoming $0\,$ at $k=1.\,$ The sign of the second term depends on $k.\,$ Let us consider two cases:

$\mathbf{\text{Case 1: }\displaystyle \frac{1}{2}\gt k\gt 0}.\,$ In this case, the second term is negative and for a fixed value f $k\,$ is a monotonically decreasing function of $m^2.\,$ Moreover, $m^2\lt k^2\,$ implies

\displaystyle \begin{align} f&\gt (5-2k)(k-1)^2+(2k-1)k^2\\ &=\frac{(4k-3)^2}{2}+\frac{1}{2}\\ &\gt\frac{1}{2}. \end{align}

$\mathbf{\text{Case 2: }\displaystyle \frac{3}{2}\gt k\ge\frac{1}{2}}.\,$ In this case, the second term is non-negative and takes value $0\,$ when $\displaystyle k=\frac{1}{2}\,$ or $m=0.\,$ Thus, $f\,$ takes minimum value of $0\,$ when $k=1\,$ andf $m=0.$

### Solution 6

From $ab+bc+ca=3,\,$ $a^2+b^2+c^2\ge 3,\,$ by the AM-GM inequality. If we use 3D polar coordinates, then the problem reduces to finding the maximum of

$\displaystyle \frac{1}{\rho^2\sin^2\theta\cos^2\varphi}+\frac{1}{\rho^2\sin^2\theta\sin^2\varphi}+\frac{1}{\rho^2\cos^2\theta},$

where, as we just observed $\rho^2\ge 3.\,$ The above function decreases as $rho\,$ grows so that it achieves its maximum on the sphere $rho^2=3\,$ - for the minimal value of $rho.$

Further, from $ab+bc+ca=3\,$ and $a^2+b^2+c^2=3\,$ it follows that $(a+b+c)^2=9,\,$ i.e., $a+b+c=3,\,$ the tangent plane to the sphere $a^2+b^2+c^2\,$ at the point $(1,1,1).\,$ This proves the required inequality as

$\displaystyle \frac{1+a^2}+\frac{1+b^2}+\frac{1+c^2}\le\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1.$

### Illustration ### Acknowledgment

Leo Giugiuc has kindly posted at the CutTheKnotMath facebook page the above problem which he coauthored with Dan Sitaru. The problem has been included at the 2016 Danubius contest. Solution 1 is by the authors; Solution 2 is by Marian Daniel Vasile; Solution 3 is by Vasile Cîrtoaje; Solution 4 is by Srinivas Vemuri; Solution 5 is by Amit Itagi; Solution 6 is by Andrea Acquaviva. The illustration is by N. N. Taleb.

A few variants of this problem have been discussed elsewhere. 