# An Inequality with Constraint XII

### Solution 1

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

$1\ge x\ge y\ge z\ge 0\;$ and $x+y+z=t.$

The required inequality translates into

$-xyzt-xyz+t(xy+yz+zx)\ge 0$

which is equivalent to

$(x+y+z)(xy+yz+zx)\ge xyz(x+y+z+1).$

The latter inequality is true because, by the AM-GM inequality, $(x+y+z)(xy+yz+zx)\ge 9xyz\;$ and the obvious $4\ge 1+x+y+z.$

Let's study the cases of equality. If $xyz\gt 0,\;$ the inequality is strict. Assume then $xyz=0.\;$ Then either $x+y+z=0\;$ or $xy+yz+zx=0.\;$ It follows that $(x,y,z)=(\alpha,0,0),\;$ $\alpha\in [0,1].\;$ In other words, the equality is achieved only when $(a,b,c,d)=(1+\alpha,1,1,1-\alpha).$

### Solution 2

Define $f(a,b,c,d)=5+abcd-(ab+bc+cd+da+ac+bd).\;$ Then

f(a,b,c,d)-f(a+b-1,1,c,d)=(1-cd)(a-1)(1-b)\ge 0\\ \begin{align} f(a+b-1,1,c,d)&=(a+b-2)(a+b+cd-3)\\ &=(a+b-2)(1-c)(1-d)\\ &\ge 0. \end{align}

### Acknowledgment

The problem and Solution 1 are by Leo Giugiuc who also communicated to me Solution 2 from the Art of Problem Solving forum.