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I spent an hour or so yesterday trying to solve the inequality $x^2 + x + 1 > 0$. Since I'd spent so long on a problem didn't seem like it should be that difficult, I decided I'd call it a day and try it again later.

I just had another look at it and this solution became immediately obvious:

$$x^2 + x + 1 > 0 \ \ \forall \ \ x \in \mathbb{R}$$

I'd justify this by stating that $x^2 > x \ \ \forall \ \ x \in \mathbb{R}$. Because of this, even if $x < 0$, the right hand side of the inequality will always be positive. Am I correct?

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actually $x^2 < x $ for $0 <x <1$. –  Hardy Mar 9 '12 at 23:15
Your claim is false $(1/2)^2< 1/2$. –  azarel Mar 9 '12 at 23:16
Didn't think about that .. –  jamesbrewr Mar 9 '12 at 23:16
It is true that $x^2 + x + 1 > 0$ for all $x \in \mathbb{R},$ but not for the resons you give. If $0 < x < 1,$ we have $x^2 < x.$ –  Geoff Robinson Mar 9 '12 at 23:16

6 Answers 6

Hint $\rm\ \ 4\:f(x)\: =\: 4\:(x^2+x+1)\ =\ (2x+1)^2 + 3\: \ge\: 3\ $ thus $\rm\:f(x)\ge 3/4$

More generally one can decide polynomial (in)equations by partitioning the real line into intervals based on the finite number of roots, and testing on each interval. Here there are no real roots, so it has constant sign, so has the sign of $\rm\:f(0) = 1.\:$

This is a special case of Collin's cylindrical algebraic decomposition algorithm, an effective form of Tarski-Seidenberg quantifier elimination for the reals. For deep generalizations, search on "semialgebraic geometry" and "o-minimal".

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Is there any special significance in choosing to multiply by 4 and add 3, or is this just a case that works out? –  jamesbrewr Mar 9 '12 at 23:25
@James It's a convenient way to complete the square, just as in the classical quadratic formula. –  Bill Dubuque Mar 9 '12 at 23:32

$$x^2 + x + 1 > 0 \Leftrightarrow x^2 + x + \frac{1}{4} > -1 + \frac{1}{4} \Leftrightarrow (x+\frac{1}{2})^2 > -\frac{3}{4} $$ And we have the desired result by the positivity of the square. (i.e. $\forall x \in \mathbb{R}: x^2 \geq 0$)

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First I will show that with small modifications, your approach can be made to work. (It is, however, simpler to use one of the methods in other answers.) At the end, I suggest an additional simple approach.

It is clear that $x^2+x+1 \ge 1$ if $x \ge 0$. There is potential trouble only when $x<$, so suppose that $x<0$. Because it is all too easy to make errors when handling negative numbers, let $w=-x$. Then $x^2+x+1=w^2-w+1$, and $w$ is positive.

If $w \ge 1$ (that is, if $x\le -1$), then $w^2 \ge w$, so $w^2 -w+1\ge 0$. (Here we used the method that you proposed.)

That method doesn't work when $0<w<1$, for then $w^2<w$. But $w^2>0$ and $w<1$, so $w^2-w>-1$, and therefore $w^2-w+1>0$.

Another way: Note that $(x-1)(x^2+x+1)=x^3-1$, and that $x-1$ and $x^3-1$ are positive if and only if $x>1$, and are negative if and only if $x<1$.

If $x>1$, then $x-1$ and $x^3-1$ are both positive, so the ratio $\frac{x^3-1}{x-1}$, that is, $x^2+x+1$, is positive.

If $x-1$ and $x^3-1$ are both negative, then again their ratio $x^2+x+1$ is positive.

And finally if $x=1$, then $x^2+x+1$ is positive.

Note that in exactly the same way, we can show that $x^n+x^{n-1}+\cdots +x+1$ is always positive if $n$ is even.

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I bring x and 1 to the other side to change the problem to $x^2 > -1 - x$.

Note that $x^2$ is always positive for real $x$, so the equality can only be false when $x < -1$.

But, if $x < -1$, $x^2$ is still positive, and so is $- x$. So we need to prove that when $x > 1$, then $x^2 > x - 1$ holds. But since $x^2 > x$ when $x > 1$ this holds.

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There's also this answer (but I'm not sure if it qualifies as a proof): Plot of $y=x^2+x+1$, notice it never touches the $x$-axis

This is a plot of $y=x^2+x+1$, notice it never touches the $x$-axis.

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It may be good to add some words to this chart to show how it solves the problem. –  Emmad Kareem Mar 10 '12 at 11:37
Graphing your data is a good way to get started to solve a problem. –  Frank Meulenaar Mar 15 '12 at 18:30

One more way to solve this (actually a different approach):

Since $1^2+1+1=3>0$ any y for which $y^2+y+1 \leq 0$ holds would yield a root of the polynomial $p(x)=x^2+x+1$ by the intermediate value theorem. Therefore you get a second root (besides $1$) of the polynomial $q(x)=p(x)(x-1)=x^3-1$. But this is a monotonically increasing function and hence has at most one zero.

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