An Inequality with Constraint

Statement

Leo Giugiuc has kindly posted a problem of proving an inequality at the CutTheKnotMath facebook page with a solution (Solution 1) by Claudia Nanuti, Diana Trailescu, Dan Sitaru and Leo Giugiuc and a reference to the AOPS forum where I found another solution (Solution 2) by a forum participant under the moniker sqing. I took the liberty of borrowing that essentially different solution. Solution 3 is by Grégoire Nicollier.

The problem has also been published earlier by Marius Stanean, Romania, without a solution, in the Mathematical Reflections, Number 1, 2013 (Problem J258).

Assume $x,y,z$ positive real numbers that satisfy $x\le 1,$ $y\le 2,$ $x+y+z=6.$ Prove that

$(x+1)(y+1)(z+1)\ge 4xyz.$

The illustration below is by Gary Davis.

Solution 1

The inequality is equivalent to $7+xy+yz+xz\ge 3xyz,\;$ or

$7+xy+(x+y)[6-(x+y)]\ge 3xy[6-(x+y)].$

We'll consider two cases:

Case 1: $y\le 1.$

Set $x+y=2s,$ xy=$p^2.$ Obviously, $0,\le p\le s\le 1.\;$ The task becomes to prove

$7-17p^2+12s-4s^2+6p^2s\ge 0.$

The function $f(s)=7-17p^2+12s-4s^2+6p^2s\;$ is strictly increasing on $[p,1],\;$ so that, on that interval,

\begin{align} \min f(s)=f(p) &= 6p^3-21p^2+12p+7\\ &= 3(1-p)(1+5p-2p^2)+4\gt 0, \end{align}

for $p\in[0,1].$

Case 2: $1\le y\le 2.$

Consider the function $g(x)=(3y-1)x^2-(19y-3y^2-6)x+7+6y-y^2\;$ on $[0,1].$ Note that, for $y\in [1,2],$ $\displaystyle\frac{19y-3y^2-6}{2(3y-1)}\gt 1.$ Thus, for $x\in [0,1],$

$\min g(x)=g(1)=2(2-y)(3-y)\gt 0.$

This completes the prove of the required inequality.

Solution 2

By the AM-GM inequality, $\displaystyle\frac{6x+3y+2z}{3}\ge\sqrt[3]{6x\cdot 3y\cdot 2z}.$ In other words,

\displaystyle\begin{align} xyz&\le\frac{1}{6^2\cdot 3^3}(6x+3y+2z)^3\\ &\le\frac{1}{6^2\cdot 3^3}[2(x+y+z)+4x+y]^3\\ &\le\frac{1}{6^2\cdot 3^3}[12+4+2]^3\\ &=6. \end{align}

To continue,

$\displaystyle\frac{(x+1)(y+1)(z+1)}{4xyz}\ge\frac{3}{4}+\sqrt[3]{\frac{3}{32xyz}}.$

And I confess to not understanding the last step. However, it does the job:

$\displaystyle \frac{3}{4}+\sqrt[3]{\frac{3}{32xyz}}\ge\frac{3}{4}+\sqrt[3]{\frac{3}{32\cdot 6}}=\frac{3}{4}+\frac{1}{4}=1.$

The best I managed to do was $\displaystyle\frac{(x+1)(y+1)(z+1)}{4xyz}\ge 0.9544\ldots.$

Solution 3

Write $\displaystyle 2\cdot\frac{u}{2}$ as an abbreviation $\displaystyle \frac{u}{2}+\frac{u}{2}$ and use it in the AM-GM ineqality for $6$ terms applied to $6 = x+y+z\;$ with nonnegative $x,$ $y,$ $z:$

$\displaystyle 6^6 = \left(x + 2\cdot\frac{y}{2} + 3\cdot\frac{z}{3}\right)^6 \ge 6^6 x \left(\frac{y}{2}\right)^2 \left(\frac{z}{3}\right)^3,$

thus

(1)

$x y^2 z^3 \le 2^2 3^3.$

Using $x\le 1,$ $y\le 2,$ and (1), one gets

(2)

$\displaystyle x^6 \left(\frac{y}{2}\right)^8 \left(\frac{z}{3}\right)^9 = \frac{(xyz)^{12}}{2^8 3^9 x^5 y^2 x y^2 z^3} \ge \left(\frac{xyz}{6}\right)^{12}.$

The AM-GM ineqalities for $2,$ $3,$ and $4$ terms give now together with (2)

$\displaystyle \left[(x+1)\left(2\cdot\frac{y}{2} + 1\right)\left(3\cdot\frac{z}{3} + 1\right)\right]^{12}\ge 2^{12} x^6 3^{12} \left(\frac{y}{2}\right)^8 4^{12} \left(\frac{z}{3}\right)^9\ge\left(24 \frac{xyz}{6}\right)^{12}.$