An Inequality with Constraint in Four Variables II


An  Inequality with Constraint in Four Variables II


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


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 $a=1+k,\;$ $b=c=1,\,$ $d=1-k,\,$ $k\in [0,1].$


Constraint inequality in four variables, illustration


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.


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