# Sladjan Stankovik's Inequality With Constraint

### Solution

Denote $a-1=x,\;$ $b-1=y,\;$ $c-1=z,\;$ and $d-1=t.\;$ We have

$x,y,z,t\ge -1,\\ x+y+z+t=0,\\ x^2+y^2+z^2+t^2=3.$

Easily, $\displaystyle xy+yz+zt+tx+xz+yt=-\frac{3}{2}.$

We need to prove that

$\displaystyle\sum_{cycl}(x+1)(y+1)(z+1)-\prod_{cycl}(x+1)\le\frac{27}{16},$

which is equivalent to $\displaystyle -xyzt\le\frac{3}{16},\;$ or $\displaystyle xyzt\ge -\frac{3}{16}.\;$ But according to the problem 4052 from Crux Mathematicorum, if $x+y+z+t=0\;$ and $xy+yz+zt+tx+xz+yt=-k,\;$ $k\gt 0,\;$ then $\displaystyle xyzt\ge -\frac{k^2}{12};\;$ equality holds for the permutations of

$\displaystyle \left(\sqrt{\frac{k}{6}},\sqrt{\frac{k}{6}},\sqrt{\frac{k}{6}},-3\sqrt{\frac{k}{6}}\right)\;\text{or}\;\left(\sqrt{-\frac{k}{6}},-\sqrt{\frac{k}{6}},-\sqrt{\frac{k}{6}},3\sqrt{\frac{k}{6}}\right).$

In our case, taking $\displaystyle k=\frac{3}{2}\;$ shows that the required inequality is true, with equality at $\displaystyle\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},\frac{3}{2}\right).\;$ This proves the original inequality, $\displaystyle abc+abd+acd+bcd-abcd\le\frac{27}{16},\;$ with the equality at $\displaystyle\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{5}{2}\right).$

### Acknowledgment

The problem and the above solution have been kindly posted at the CutTheKnotMath facebook page by Leo Giugiuc.

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