# Elegant Minimum of a Function

The problem below and its solution are due to Leo Giugiuc; it was offered at the "Danubius" contest.

Let $P(x)=(x^{2}+x+8)^{2}+(2x-2)^{2}.$ Define $f:\;\mathbb{R}\rightarrow\mathbb{R}$ by $f(x)=x^{2}+x+\sqrt{P(x)}.$

Find the minimum of $f(x).$

Solution

Let $P(x)=(x^{2}+x+8)^{2}+(2x-2)^{2}.$ Define $f:\;\mathbb{R}\rightarrow\mathbb{R}$ by $f(x)=x^{2}+x+\sqrt{P(x)}.$

Find the minimum of $f(x).$

### Solution

Observe that, for all $x\in\mathbb{R},$ $\sqrt{P(x)}\ge |x^{2}+x+8|\ge -(x^{2}+x+8);$ and, since $\sqrt{P(x)}\gt 0,$ $\sqrt{P(x)}\gt -(x^{2}+x+8),$ for all $x\in\mathbb{R}.$ From here, trivially, the sought minimum value is not less than $-8.$

Interestingly, $(x^{2}+x+8)^{2}+(2x-2)^{2}$ is not the only way to represent $P(x)$ in the form $(x^{2}+x+a)^{2}+(bx+c)^{2}.$ Let's see what we can find assuming $(x^{2}+x+8)^{2}+(2x-2)^{2}=(x^{2}+x+a)^{2}+(bx+c)^{2}.$ First, we may simplify that to

(1)

$(bx+c)^{2}=(20-2a)x^{2}+2(4-a)x+(68-a^{2}).$

The fact that the left-hand side has the discriminant of $0$ implies that this is also true for the right-hand side. This leads to an equation for $a:$

(2)

$2a^{3}-21a^{2}-128a+1344=0.$

From the original definition of $P(x)$ we know that $a=8$ is one of the solutions to (2), such that (2) becomes

(3)

$(a-8)(2a^{2}-5a-168)=0,$

with two additional roots: $a=-8$ and $\displaystyle a=\frac{21}{2}.$ For the latter, the right-hand side of (1) does not amount to a complete square. But, with $a=-8,$ (1) implies $(bx+c)^{2}=(6x+2)^{2},$ i.e.,

$P(x)=(x^2+x-8)^{2}+(6x+2)^{2}.$

So, as before, $\sqrt{P(x)}\ge |x^2+x-8|\ge -(x^2+x-8),$ implying $x^{2}+x+\sqrt{P(x)}\ge 8.$ This inequality may be said to be derived under the assumption $(6x+2)^{2}=0.$ Thus, it is natural to check what's happening for $\displaystyle -\frac{1}{3}.$ A quick verification shows that $\displaystyle P(-\frac{1}{3})=8$ and, therefore, $\min (P(x))$ is exactly $8:$ $\displaystyle P(x)\ge P(-\frac{1}{3})=8.$