# Minimum value of given expression

What is the minimum value of the $$\frac {x^2 + x + 1 } {x^2 - x + 1 } \ ?$$

I have solved by equating it to m and then discriminant greater than or equal to zero and got the answer, but can algebraic manipulation is possible

• You can find the minimum value of a function by setting the derivative of the function to zero and solving the resulting expression for $x$. – Daryl Jul 25 '12 at 4:42
• Also, you can simplify slightly by noting that $\frac {x^2 + x + 1 } {x^2 - x + 1 } = 1+\frac {2x } {x^2 - x + 1 }$. – copper.hat Jul 25 '12 at 4:44
• @Daryl: The question is labeled [algebra-precalculus], which would seem to mean we should check for non-calculus solutions. – Arturo Magidin Jul 25 '12 at 4:44

let y=$\frac{x^2+x+1}{x^2-x+1}$

=>$x^2(y-1)-x(y+1)+(y-1)=0$

As x is real, the discriminant= $(y+1)^2-4(y-1)^2≥0$

=>$(y-3)(y-\frac{1}{3})≤0$

=>$\frac{1}{3}≤y≤3$

• +1 Cute technique; must have missed that with my 'new maths'... – copper.hat Jul 25 '12 at 6:23
• To get from $1/3\leq y \leq 3$ to asserting that the minimum value is actually $1/3$, you need to demonstrate that $1/3$ is actually a possible value for $y$. In other words, you must show that there is a value of $x$ that gives this. Of course, there is, it's $x=-1$, but you really need to give this as part of your solution. – user22805 Jul 25 '12 at 9:06
• +2n+1 :D This technique is so easy to understand and suitable <3 – William Phoenix Sep 5 '16 at 9:20

Here is an 'algebra solution':

$\frac {x^2 + x + 1 } {x^2 - x + 1 } = \frac{(x+1)^2-(x+1)+1}{(x+1)^2-3(x+1)+3} = \frac{1}{3} + \frac{2}{3} \frac{(x+1)^2}{x^2-x+1} = \frac{1}{3} + \frac{2}{3} \frac{(x+1)^2}{(x-\frac{1}{2})^2+\frac{3}{4}}$.

Since the last term is greater than zero when $x\neq -1$, we see that the minimum is

$\frac{1}{3}$.

Of course, this is cheating since I know the answer from J.D.'s solution and this suggests the way to expand the expression.

Hint: Take the derivative w.r.t $x$ and equate it with zero, you get: $$\frac{d}{dx} \frac {x^2 + x + 1 } {x^2 - x + 1 } = - \frac {2(x^2 - 1)} {(x^2 - x + 1)^2} = 0.$$ So $x = \pm 1$ at the extrema. Test both for minimum.

Edit: this tutorial page might be helpful.

• But again, this is not algebra as possible. – user2468 Jul 25 '12 at 4:46

For $$x\geq0$$ we have $$\frac{x^2+x+1}{x^2-x+1}=1+\frac{2x}{x^2-x+1}\geq1.$$ For $$x<0$$ by AM-GM we obtain: $$\frac{x^2+x+1}{x^2-x+1}=1+\frac{2x}{x^2-x+1}=1+\frac{2}{x+\frac{1}{x}-1}=$$ $$=1-\frac{2}{-x+\frac{1}{-x}+1}\geq1-\frac{2}{2\sqrt{-x\cdot\frac{1}{-x}}+1}=\frac{1}{3}.$$ The equality occurs for $$-x=\frac{1}{-x}$$ or for $$x=-1,$$ which says that we got a minimal value.

By the same way we can get a maximal value if you want.