# An Inequality with Just Two Variables VI

### Solution 1

WLOG, $y\ge x.\,$ Let $x=-1+b,\,$ $b\in [0,2].\,$ If $y=x+a=-1+a+b\,$ then $a+b\le 2\,$ because $y\le 1.$

The inequality becomes $ab(a+b)\le 2.\,$ But $ab\le\displaystyle \frac{(a+b)^2}{4},\,$ so that $ab(a+b)\le\displaystyle \frac{(a+b)^3}{4}\le \frac{2^3}{4}=2.\,$

Equality attained for $a=b=1,\,$ i.e., $(x,y)=(0,1)\,$ or $(x,y)=(1,0).$

### Solution 2

Set $1+x=a,\,1+y=b.\,$ Then $0\le a,b\le 2,\,$ and we need to show $|a-b|ab\le 2.\,$

WLOG, $a\ge b,\,$ converting inequality into $(a-b)ab\le 2.\,$ By the AM-GM inequality, $\sqrt{(a-b)b}\le\displaystyle \frac{(a-b)+b}{2}=\frac{a}{2},\,$ such that $\displaystyle (a-b)ab\le\frac{a^3}{4}\le\frac{8}{4}=2.\,$ Equality is attained at $a=2\,$ and $a-b=b,\,$ i.e., $b=1.\,$ So the general case of equality is $x=1,\,y=0\,$ or $x=0,\,y=1.$

### Solution 3

Since $x,y\in [-1,1],\,$ we can define $x=\cos 2a;\,$ $y=\cos 2b,\,$ where $a,b\in [0,\pi].\,$ We have

\displaystyle \begin{align} LHS &= \cos 2a-\cos 2b|(\cos 2a+1)(\cos 2b+1)\\ &= 4\cdot|(2\cos^2a-1)-(2\cos^2b-1)|\cos^2a\,\cos^2b\\ &=8\cos^2a\,\cos^2b|\cos^2a-\cos^2b|\\ &\le 8\cos^2a\,\cdot (1-\cos^2a),\,\text{(because}\,|\cos t|\le 1,\,\forall t)\\ &\le 8\frac{\cos^2a+1-\cos^2a}{4}=2. \end{align}

Equality holds at $2\cos^2=1,\,\cos^2b=1,\,$ i.e., $\cos 2a=0\,$ and $\cos 2b=1,\,$ or $(x,y)=(0,1)\,$ or $(x,y)=(1,0).$

### Solution 4

The square region of interest can be split into two right triangles (along the diagonal $x=y).\,$ We prove the inequality in the lower triangle. The proof in the upper triangle follows from the symmetry in $x\,$ and $y.$

The lower triangle has vertices $(-1,-1),\,$ $(1,-1),\,$ and $(1,1).\,$ Let $p=x+1\,$ and $q=y+1.\,$ Thus, we need to prove $pq(p-q)\le 2\,$ in the triangle with vertices $(0,0),\,$ $(2,0),\,$ and $(2,2)\,$ in the $PQ\,$ plane.

Consider the family of lines $p-q=t,\,$ with $t\in [0,2].\,$ These are parallel lines with varying intercepts and cover the entire triangle. For a fixed value of $t,\,$ the segment of the line in (and on) the triangle is given by $p\in [t,2].$

\displaystyle \begin{align} LHS &= pq(p-q)=p(p-t)t\\ &=t\left[\left(p-\frac{t}{2}\right)^2-\frac{t^2}{4}\right] \end{align}

Thus, for a fixed $t,\,$ the maximum value of LHS is

$\displaystyle t\left[\left(2-\frac{t}{2}\right)^2-\frac{t^2}{4}\right]=t(4-2t)=2-2(t-1)^2.$

This is a quadratic function that takes maximum value at $t=1.\,$ The maximum value is $2.$

One more

### Acknowledgment

The problem has been posted but Nguyen Viet Hung at the mathematical inequalities facebook group and kindly reposted by Leo Giugiuc to the CutTheKnotMath facebook page. Solution 1 is by Imad Zak (Lebanon); Solution 2 is by Leo Giugiuc (Romania); Solution 3 is by Pahm Quy (Vietnam); Solution 4 is by Amit Itagi (USA). The illustration is by N. N. Taleb.

Inequalities in Two Variables