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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

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    You can find the minimum value of a function by setting the derivative of the function to zero and solving the resulting expression for $x$.2012-07-25
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    Also, you can simplify slightly by noting that $\frac {x^2 + x + 1 } {x^2 - x + 1 } = 1+\frac {2x } {x^2 - x + 1 }$.2012-07-25
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    @Daryl: The question is labeled [algebra-precalculus], which would seem to mean we should check for non-calculus solutions.2012-07-25

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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= $(y+1)^2-4(y-1)^2≥0$

=>$(y-3)(y-\frac{1}{3})≤0$

=>$\frac{1}{3}≤y≤3$

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    +1 Cute technique; must have missed that with my 'new maths'...2012-07-25
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    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.2012-07-25
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    +2n+1 :D This technique is so easy to understand and suitable <32016-09-05
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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.

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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.

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    But again, this is not algebra as possible.2012-07-25
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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.