# Two Constraints, One Inequality by Qing Song

### Problem

### Solution 1

Clearly, none of $a,b,c$ is zero. Let $P(x)=(x-a)(x-b)(x-c)$ and $\displaystyle f(x)=\frac{P(x)}{x},$ the latter defined on $(-\infty,0)\cup (0,\infty),$ both with the roots $a,b,c.$

By Viète's theorem, $P(x)=x^3+2x^2+qx+4,$ for some real $q.$

Observe that $\displaystyle f'(x)=\frac{2(x-1)(x^2+2x+2)}{x^2},$ implying that $f'(x)\lt 0$ for $x\lt 0,$ making $f$ strictly decreasing on $(-\infty,0).$ With $f(-\infty)=\infty$ and $f(0^{-})=-\infty,$ $f$ has a unique root on $(-\infty,0)$ such that on $(0,\infty)$ it has two roots separated by $x=1,$ hence distinct. Since $f(0^+)=f(\infty)=\infty,$ $f(1)\lt 0.$ (For the record, also $P(1)\lt 0.)$

So, WLOG, assume $a,b\gt 0$ and $c\lt 0.$ Then $c=-2-a-b,$ implying $|c|=a+b+2.$ The required inequality reduces to $a+b\ge 2.$ Now, verifiably, $P(-4)=-4P(1)\gt 0$ and, subsequently, $c\le -4.$

Finally, $-2=a+b+c\le a+b-4,$ so that $a+b\ge 2,$ and the required inequality follows.

Equality is attain for $a=b=1$ and $c=-4$ and permutations.

### Solution 2

If all $a,b,c$ are negative (not one could be zero), then $2=|a|+|b|+|c|\ge 3\sqrt[3]{|abc|},$ implying that $|abc|\le 1$ in contradiction to $abc=-4.$ Hence, two are positive and one is negative. Assume $a,b\gt 0$ and $c\lt 0.$

If $a+b\lt 2$ then $ab\lt 1$ so that $c\le -4.$ But from $a+b+c=-2$ it follows that $a+b\gt 2.$ A contradiction. Hence, $a+b\ge 2,$ implying $c\le -4.$ And, finally, $|a|+|b|+|c|\ge 6.$

### Solution 3

All three $a,b,c$ could not be negative. Indeed, using $x,y,z$ as their absolute values, on one hand $\displaystyle \frac{x+y+z}{3}=\frac{2}{3}\lt 1,$ while, on the other hand, $(xyz)^{\frac{1}{3}}=4^{\frac{1}{3}}\gt 1.$

Continue as above.

### Solution 4

$x+y+z=2$ and $xyz=4$

And let $F(x,y,z)=|x|+|y|+|z|.$ Note that, from the AM-GM inequality, all three $x,y,z$ could not be positive. WLOG, assume $x\gt 0 \gt, y,z.$ Then $F(x,y,z)=x-y-z=2x-2.$

Claim: $x\ge 4.$

Assume to the contrary that $x\lt 4.$ then $0 \lt -y-z=x-2\lt 2$ and

$yz=(-y)(-z)\le (-y)(2+y)\le 1,$

for all $0\lt -y\le 2.$ Multiplying this by $x$ gives $xyz\le x\lt 4,$ contrary to constraint. Thus, $x\ge 4$ and $F=2(x-1)\ge 6.$

### Acknowledgment

This problem by Qing Song (Bejing) was kindly posted at the CutTheKnotMath facebook page by Leo Giugiuc, along with a solution of his Solution 1. Solution 2 is by Christopher D. Long; Solution 3 is by Dan Davies; Solution 4 is by Sam Walters.

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