# Three Variables, Three Constraints, Two Inequalities (Only One to Prove) - by Leo Giugiuc

### Solution 1

At least one of $a,b,c\le 0.$ Let it be $c:$ $-1\le c\le 0.$ Then, on squaring, $0\le c^2\le 1.$

From $a+b+c=0,$ $(a+b)^2=c^2.$ Further, from $a^2+b^2+c^2\ge 2,$

\begin{align} 2+2ab &\le a^2+b^2+2ab+c^2=(a+b)^2+c^2\\ &=2c^2\le 2. \end{align}

But $2c^2\le 2,$ such that $2+2ab\le 2,$ i.e., $ab\le 0.$ Now, since also $c\le 0,$ $abc\ge 0.$

### Solution 2

From $(a+b+c)^2=0$ and $2\le a^2+b^2+c^2=-2(ab+bc+ca),$ it follows that $ab+bc+ca\le -1$ which is equivalent to $ab+c(a+b)\le -1,$ i.e., $ab-c^2\le -1$ and, subsequently, $ab\le c^2-1.$ WLOG, $-1\le c\le 0,$ implying $c^2-1\le 0.$ Thus, $ab\le 0$ and, since $c\le 0,$ $abc\ge 0.$

### Solution 3

Let $ab+bc+ca=-q.$ Then $q\ge 1.$ But $(-1-a)(-1-b)(-1-c)\le 0$ implies $q-1\le abc$ and, since $0\le q-1,$ $abc\ge 0.$

Equality is attained for $(a,b,c)=(-1,0,1)$ and permutations.

### Solution 4

1. $a,b,c$ can't all be negative or all positive.

2. Suffice it to show that only one of them can't be negative.

3. Let $c\lt 0,$ making $a+b\lt 1,$ such that $a^2+b^2+c^2\ge 2$ implies $a^2+b^2+ab-1\ge 0.$

4. The quadratic equation in 'a' can't have a positive discriminant, so $\displaystyle b\ge\frac{1}{\sqrt{3}}$ and the same for $a:$ $\displaystyle a\ge\frac{1}{\sqrt{3}}$.

5. It follows that $\displaystyle a+b\ge\frac{2}{\sqrt{3}}\gt 1$ in contradiction with $a+b\lt 1.$

### Acknowledgment

The problem, with a solution (Solution 3), was kindly posted by Leo Giugiuc at the CutTheKnotMath facebook page. Solution 1 is by Marian Cucoaneş; Solution 2 is by Nguyễn Ngọc Tú; Solution 4 is by Amitabh Bachchan.