# Sladjan Stankovik's Inequality In Four Variables

### Solution 1

WLOG, $a+b+c+d=4.\,$ Then $\displaystyle \sum_{all}ab=6(1-t^2),\,$ $\displaystyle \sum_{cycl}=4(1+3t^2)\,$ and $\displaystyle \sum_{cycl}a^3=4(18t^2+3s-2),\,$ where $0\le t\le 1\,$ and $4s=\displaystyle \sum_{cycl}abcd.$

p>We shall consider three cases:

Case 1: $\displaystyle \mathbf{0\le t\le\frac{1}{3}}.$

According to Vo Quoc Ba Can, $-2t^3-3t^2+1\le s.\,$ Suffice it to show that

\begin{align} &8(1+3t^2)\le 3(-6t^3+9t^2+1)+6(1-t^2)\;\Leftrightarrow\\ &(1-3t)(6t^2+3t+1)\ge 0, \end{align}

which is true.

Case 2: $\displaystyle \mathbf{\frac{1}{3} \le t\lt\frac{1}{\sqrt{3}}}.$

There exist $\alpha\ge\beta\gt 0\,$ such that $2\alpha+\beta=4\,$ and $2\alpha^2+\beta^2=t(1+3t^2).\,$ With these $\alpha\,$ and $\beta,\,$ $\displaystyle \sum_{cycl}a^3\ge 2\alpha^3+\beta^3.\,$ Suffice it to show that

\begin{align} &2(2\alpha^2+\beta^2\le 3\frac{2\alpha^3+\beta^3}{2\alpha+\beta}+\alpha(\alpha+2\beta)\;\Leftrightarrow\\ &(\alpha-\beta)^2\ge 0, \end{align}

which is true.

Case 3: $\displaystyle \mathbf{\frac{1}{\sqrt{3}}} \le 1.$

In this case, $s\ge 0,\,$ hence, suffice it to show that $3t^2-1\ge 0,\,$ which is true.

Equality occurs at $(a,a,a,0)\,$ and $(a,a,0,0)\,$ and permutations.

### Solution 2

Let $a+b=s_1,\,$ $c+d=s_2,\,$ $ab=p_1,\,$ $cd=p_2.\,$ WLOG, we may assume $a+b+c+d=1.\,$ Then

\displaystyle \begin{align} &a^2+b^2=s_1^2-2p_1,\;c^2+d^2=s_2^2-2p_2,\;s_1+s_2=1,\\ &a^3+b^3=s_1^3-3p_1s_1,\;c^3+d^3=s_2^3-3p_2s_2,\\ &ab+ac+ad+bc+bd+cd=ab+cd+(a+b)(c+d)\\ &\qquad\qquad\qquad\qquad\qquad=p_1+p_2+s_1s_2\\ &2(a_1^2-2p_1+s_2^2-2p_2-3(s_1^3-3p_1s_1+s_2^2-3p_2s_2)\le p_1+p_2+s_1s_2. \end{align}

WLOG, we may assume that $s_1\le s_2\,$ so that $\displaystyle s_1\le\frac{1}{2}.\,$ Define function $\displaystyle f:\,\left[0,\frac{s_1^2}{4}\right]\to\mathbb{R}\,$ with

$f(p)=2(a_1^2-2p+s_2^2-2p_2)-3(s_1^3-3ps_1+s_2^2-3p_2s_2)-(p+p_2+s_1s_2).$

Then $\displaystyle f'(p)=-4+9s_1-1=-5+9s_1\le -5+\frac{9}{2}\lt 0\,$ so that $f\,$ is decreasing:

\begin{align} f(p)\;&\le f(0)=2(s_1^2+s_2^2-2p_2)-3(s_1^3+s_2^3-3p_2s_2)-(p_2+s_1s_2)\\ &=2(1-s_1s_2-2p_2)-3(1-3s_1s_2-3p_2s_2)-(p_2+s_1s_2)\\ &=4s_1s_2+p_2(9s_2-5)-1\\ &=4(1-s_2)s_2+p_2(9s_2-5)-1. \end{align}

Let $g(p)=4(1-s_2)s_2+p(9s_2-5)-1.\,$ $g'(p)=9s_2-5.\,$

If $\displaystyle s_2\le\frac{5}{9}\,$ then $g\,$ is decreasing,

$g(p)\le g(0)=4(1-s_2)s_2-1=-(2s_2-1)^2\le 0.$

In particular $g(p_2)\le 0.\,$

If $\displaystyle s_2\ge\frac{5}{9}\,$ then $g\,$ is increasing,

\displaystyle \begin{align} g(p)&\le g\left(\frac{s_2^2}{4}\right)=4(1-s_2)s_2+\frac{s_2^2}{4}(9s_2-5)-1\\ &=\frac{9s_2^3-21s_2^2+16s_2-4}{4}\\ &=\frac{(s_2-1)(9s_2^2-12s_2+4)}{4}\\ &=\frac{(s_2-1)(3s_2-2)^2}{4}\le 0, \end{align}

because $s_2\le 1.\,$ It follows that in this case, too, $g(p_2)\le 0.$

We conclude that $f(p_1)\le f(0)=g(p_2)\le 0,\,$ which proves the required inequality.

### Solution 3

The inequality being homogeneous, WLOG assume $a+b+c+d=1$. The inequality can be simplified using the following result:

$(a+b+c+d)(a+b+c+d-1)=0 \Rightarrow \sum_{cyc} a^2+2\sum_{all}ab-1=0,$

to

$5(a^2+b^2+c^2+d^2)-6(a^3+b^3+c^3+d^3)\leq 1.$

The Lagrangian for finding the stationary points of the LHS is

$L=5(a^2+b^2+c^2+d^2)-6(a^3+b^3+c^3+d^3)-\lambda(a+b+c+d-1).$

The partial derivatives give in addition to the original constraint, a set of equations of the form: $5a_i-9a_i^2=5a_j-9a_j^2$, where $a_i,a_j\in\{a,b,c,d\}$ and $a_i\neq a_j$. The solutions are $a_i=a_j$ and $a_i=5/9-a_j$.

Thus, at the stationary points, $\{a,b,c,d\}$ gets partitioned into two equivalence classes such that the variables in the same class have the same value and two variables from the different classes sum to $5/9$. Thus, the original constraint can be written as $Nx+(4-N)(5/9-x)=1$ where $N\in\{1,2,4\}$, $N$ variables equal $x$ and $4-N$ variables equal $5/9-x$. The only possible solutions are $(N=1,x=1/3)$ and $(N=4,x=1/4)$.

For the first case,

$\displaystyle LHS=5\left[\left(\frac{1}{3}\right)^2+3\left(\frac{2}{9}\right)^2\right] -6\left[\left(\frac{1}{3}\right)^3+3\left(\frac{2}{9}\right)^3\right]=\frac{71}{81}\leq 1.$

For the second case,

$LHS=4\left[5\left(\frac{1}{4}\right)^2-6\left(\frac{1}{4}\right)^3\right]=\frac{7}{8}\leq 1.$

### Acknowledgment

The problem and his solution (Solution 1) have been kindly communicated to me by Leo Giugiuc. The problem has been originally posted at the mathematical inequalities facebook group where I found Marian Dinca's solution (Solution 2). Solution 3 is by Amit Itagi.