An Inequality with Constraint in Four Variables II

Solution

There are $y,z,t\in [0,1]\,$ such that $b=1-y,\;$ $c=1-z,\;$ $d=1-t,\,$ $a=1+y+z+t.\,$ The inequality to prove becomes

\begin{align} (1-y)^3&+(1-z)^3+(1-t)^3+(1+y+z+t)^3\\ &\qquad\qquad +6(1-y)(1-z)(1-t)(1+y+z+t)-10\\ &=9(y^2z+yz^2+y^2t+yt^2+z^2t+zt^2)\\ &\qquad\qquad +6yzt(1-y)+6yzt(1-z)+6yzt(1-t)\\ &\ge 0. \end{align}

As to the equality, all the terms in

$9(y^2z+yz^2+y^2t+yt^2+z^2t+zt^2)+6yzt(1-y)+6yzt(1-z)+6yzt(1-t)$

are non-negative. For there to be an equality, all of them should vanish. The expression in the first pair of parentheses is zero provided any two of $y,z,t\,$ vanish. The other three terms vanish automatically. When two of $y,z,t\,$ vanish, the third may be pretty much arbitrary, i.e., within $[0,1].],$ Thus, in terms of $a,b,c,d,\,$ equality is achieved when, two of $b,c,d\,$ are $1\,$ and one is in $[0,1].\,$ Given their ordering, the only possibility is $a=1+k,\;$ $b=c=1,\,$ $d=1-k,\,$ $k\in [0,1].$

Acknowledgment

The problem and the Solution have been kindly posted by Leo Giugiuc at the CutTheKnotMath facebook page. The solution is by Marian Cucoanes. Illustration by Nassim Nicholas Taleb.