# Proving $x^2+x+1\gt0$

I was doing a question recently, and it came down to proving that $x^2+x+1\gt0$. There are of course many different methods for proving it, and I want to ask the people here for as many ways as you can think of.

My methods:

1. $x^2+x+1=(x+\frac12)^2+\frac34$, which is always greater than $0$.

2. Let it be $0$ for some $x=k$. Then $x^2+x+1=0$ has a real solution. But since $1^2\not\gt4$, this has no real solution. Therefore it is more than $0$.

• $15$ answers, is this a record ? Jun 28 '18 at 11:12
• @Peter Probably not. Jun 28 '18 at 11:14
• note also that completing the square gives us both the local minima ($-\frac12$), and the corresponding $y$-value ($\frac34$).
– JMP
Jun 29 '18 at 12:25

We note that:

$$x^2+x+1=\frac{x^3-1}{x-1}$$

and the sign of the RHS numerator and denominator are always equal, except for $x=1$.

We handle $x=1$ separately, but this is trivial on the LHS.

• Nice! - This could also be interpreted as the slope of the line from $(1, f(1))$ to $(x, f(x))$ for the increasing function $f(x) = x^3$. Jun 28 '18 at 8:41
• The third proof of this inequality that I have enjoyed immensely so far. It is noteworthy that these three involve either factoring some expression or completing some square after a clever scaling. Jun 28 '18 at 10:03

Short trivial proof:

Since this is a quadratic equation, and the leading coefficient is $+1$, we have

$$\Delta < 0$$

Whence the equation is always strictly positive (that is, it's always $>0$).

• I think it's worth to note that we need the leading coefficient to be positive too, even if it's trivial. Jun 28 '18 at 7:56
• @Botond You're right! Jun 28 '18 at 8:00

There is also the following way.

For $x\geq-1$ we obtain $$x^2+x+1=x^2+(x+1)>0$$ and for $x<-1$ we obtain $$x^2+x+1=x(x+1)+1>0+1>0.$$

• This is also one of the most elegant proofs in reply to this OP. However, that ugly trick with the $0$ on the LHS of your last inequality is not needed. It suffices to note that in this case $x+1<0.$ Jun 28 '18 at 10:44
• @Allawonder The last inequality is true for $x<-1$. See better my post. Jun 28 '18 at 10:46
• Sorry. That was not what I meant. I have edited my comment. See above again. Jun 28 '18 at 10:51
• @Allawonder Yes, in this case $x(x+1)>0$, which gives $x(x+1)+1>0.$ If you wish you can fix my post. Jun 28 '18 at 10:55

If $x$ is positive, then $x^2+x+1$ is clearly positive.

If $x$ is negative then $x^2-x+1$ is certainly positive. Now $$(x^2+x+1)(x^2-x+1)=x^4+x^2+1$$ is certainly positive, so $x^2+x+1$ must also be positive in this case.

If $x$ is zero then $x^2+x+1=1\gt 0$

• This is a particularly pleasant proof. But from the product, you could simply have used the fact that if $ab>0$, then either $a,b>0$ or $a,b<0$ to conclude that in either case (whether $x>0$ or $x<0$) one of the factors is always positive. This settles it very elegantly. Indeed, that was how I interpreted it. PS. Of course, it is positive when $x=0.$ Jun 28 '18 at 9:49
• @Allawonder Indeed - good observation - that avoids cases. Jun 28 '18 at 9:51

I don't know why you need multiple proofs for a simple result, but here are two overkilling solutions. The first one utilizes some knowledge in linear algebra. The second one uses Euclidean geometry along with some trigonometry.

Consider the matrix $\mathbf{B}:=\begin{bmatrix}1&\frac12\\\frac12&1\end{bmatrix}$. Being a real symmetric $2$-by-$2$ matrix, $\mathbf{B}$ has two real eigenvalues, which are $\frac{3}{2}$ and $\frac{1}{2}$. As both eigenvalues are positive, $\mathbf{B}$ is a positive-definite matrix, whence it induces a positive-definite symmetric bilinear form $\langle\_,\_\rangle:\mathbb{R}^2\times\mathbb{R}^2\to\mathbb{R}$ sending a pair $(\mathbf{u},\mathbf{v})$ of $2$-by-$1$ column vectors in $\mathbb{R}^2$ to $$\langle \mathbf{u},\mathbf{v}\rangle:=\mathbf{u}^\top\,\mathbf{B}\,\mathbf{v}\,.$$ That is, $$\langle \mathbf{u},\mathbf{u}\rangle \geq 0\text{ for all }\mathbf{u}\in\mathbb{R}^2\,,$$ and the inequality becomes an equality iff $\mathbf{u}$ is the zero vector. In particular, when $\mathbf{u}=(x,1)$, where $x$ is an arbitrary real number, we get $\mathbf{u}\neq \boldsymbol{0}$, whence $$x^2+x+1=\langle\mathbf{u},\mathbf{u}\rangle>0\,,$$ as desired.

Alternatively, consider three points in $\mathbb{R}^2$: the origin $O=(0,0)$, the point $A=(1,0)$, and the point $B=\left(-\frac{x}{2},\frac{\sqrt{3}x}{2}\right)$. Note that $\angle AOB=\frac{2\pi}{3}$ for $x>0$, and $\angle {AOB}=\frac{\pi}{3}$ for $x<0$. Using the Law of Cosine, you get $$AB^2=x^2+x+1\,,$$ whence $x^2+x+1>0$, noting that $A\neq B$ for any value of $x$. (The case $x=0$ can be checked separately, but then $x^2+x+1=AB^2=1>0$ still holds.)

Here a rather geometric way: $$y = x^2+x+1 = x(x+1) + 1$$ So, $y = x^2+x+1$ is the parabola $y=x(x+1)$ shiftet by $1$ upwards.

$y=x(x+1)$ has its vertex at $x_V = -\frac{1}{2} \Rightarrow y_V = -\frac{1}{4}$

So, the vertex of $y= x(x+1) + 1$ is also at $x_V = -\frac{1}{2}$ with a minimum value of $y_{min}= -\frac{1}{4}+1 = \frac{3}{4}>0$

The desired inequality is a convex combination of two (weak) inequalities:

$$(x+1)^2 = x^2 + 2x + 1 \ge 0, \quad \text{equality iff }\; x = -1;$$ $$(x-1)^2 = x^2 - 2x + 1 \ge 0, \quad \text{equality iff }\; x = 1.$$

Now multiply the first inequality by $3/4$, multiply the second inequality by $1/4$, and add the two resulting inequalities. We get

$$x^2 +x + 1 > 0.$$

Incidentally, the inequality $x^2 + cax + a^2 \ge 0$ holds for every $c \in [-2, 2]$ and $a \in \mathbb R$, for the same reason.

Actually your first method is efficient enough, but if you want more here you go. Let $$f(x) = x^2+x+1$$ One has $f(-\frac 1 2)>0$. Moreover $f'(x) = 2x+1\geq 0$ for all $x\in[-\frac 1 2,\infty)$. Hence $f(x) >0$ for all $x\in[-\frac 1 2,\infty)$. Since $f$ is symmetric around $x=- \frac 1 2$, we conclude $f(x) >0$ for all $x\in\mathbb R$.

Remark. This method might work for proving inequalities for general differentiable $f$, however in this case it is just an overkill.

You can complete the square two ways - if we multiply by $4$ first we get: $$4(x^2+x+1)=(2x+1)^2+3=3x^2+(x+2)^2$$

The second one of these requires an extra step to note that it is never zero - $x^2$ and $(x+2)^2$ can never both be zero at the same time.

• Short and straightforward. You did not even need to expand. Completing the square after multiplying by $4$ does it in one stroke. Great! Jun 28 '18 at 9:58

$$x^2+x+1\geq x^2-2|x|+1=(|x|-1)^2\geq 0$$

(and first inequality is strict for $|x|=1$)

Claim $$x^2+x+1>0$$

Proof

It is equivalent to prove by multiplying both sides by $x-1$

\begin{cases} x^3-1<0\iff x^3<1\iff x<1, & \text{if $x<1$} \\ x^3-1>0\iff x^3>1 \iff x>1, & \text{if $x>1$} \\ x^2+x+1=3>0, &\text{if $x=1$} \end{cases}

• Very nice solution Jun 28 '18 at 10:09

One method is to find the vertex. The x-coordinate of the vertex must be equal to -b/(2a) = -1/2. Plugging this back into the function, we get that the vertex is equal to (-1/2, 3/4).

Now that we know the vertex's y-coordinate is greater than zero, and that the parabola must be pointed up (a>0), to yield a conclusion that the parabola must always be positive. The minimum value has a y-value greater than zero, and all other y-values on the function must also be greater than zero.

Using the inequality between arithmetic and geometric mean: $$(x^2 + 1) + x \ge 2\sqrt{x^2 \cdot 1} + x = 2 |x| + x \ge |x| \ge 0.$$ Equality cannot hold because $x^2 =1$ and $x = 0$ are not simultaneously true.

• From what I know of the AM-GM, it should be $x^2+x+1\geq3x$. Jun 28 '18 at 8:19
• @MalayTheDynamo: AM-GM can only be applied to non-negative numbers, so that would work only for $x \ge 0$. Jun 28 '18 at 8:21

Correct me if wrong:

Obvious for $x\ge 0.$

For $x <0$ consider $y=-x$, and

$y^2-y +1$ for $y>0$.

Hence: for $y>0$:

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

Let $f(x)=x^2+x+1$.

Then:

$$\lim_\limits{x\to\pm\infty}f(x)=\infty$$

Also the (now) minima is at $f'(x)=0$, i.e. $2x+1=0$, i.e. at $x=-\frac12$.

As $f(-\frac12)=\frac34\gt0$, we have $f(x)\gt0\;\;\forall x\in\mathbb{R}$.

Assume to the contary that $$f\left(x\right)=x^2+x+1=-m<0$$ for some $m>0$

$$x^2+x+1+m=0$$

$$x^2+x+\left(1+m\right)=0$$

Then we have $$\Delta=b^2-4ac=1-4\left(1+m\right)$$

$$=-3-3m=-3\left(1+m\right)<0$$

Hence there exist no such $x$ for which $$f\left(x\right)=x^2+x+1$$ take negative values

Case 1: $x\ge 0 \Rightarrow x^2+x+1\ge 0^2+0+1>0$.

Case 2: $x<0 \Rightarrow x^2+x+1>0 \iff \frac{x^2+x+1}{x}<0 \iff x+\frac 1x+1\overbrace{\le}^{AM-GM} -2+1<0$.

I am adding this answer, which though is similar to, but may be better than, my previous one.

Let $f(x)=x^2+x+1$ define a function $f:\mathbf R\to \mathbf R.$ Then $f'(x)=2x+1.$ Thus $f''(x)=2>0\,\,\forall x.$ Therefore $f$ is convex and has its minimum value at $x=-1/2.$ Because of this, it decreases with $x$ in $(-\infty,-1/2)$ and increases with $x$ in $[-1/2,\infty).$ Adding to this the fact that $f(-1/2)=3/4>0,$ we deduce that $f(x)>0$ for every real $x$ since $f$ is everywhere continuous -- and in particular at $x=-1/2.$

Consider proving:

$$x^2+1\gt-x$$

If $x$ is positive, because the LHS is always positive, this is true.

If $x$ is negative, make the transform $X=-x$, and so we have:

$$X^2+1\gt X$$

Dividing by the (positive) $X$ gives:

$$X+\frac1X\gt 1$$

which is true for all $X$ regardless of greater than/less than $1$, due to the reciprocal (in fact the inequality is greater than or equal to 2).

In fact, this last fact can be used to prove straight from:

$$x+1+\frac1x\lt0$$

with a negative $x$.

Note that $$\mathrm{x}^2\ge 0$$ and $$1\gt 0.$$

Case 1. If $$x\ge 0$$ then $$\mathrm{x}^2+x+1\gt 0.$$

Case 2. Assume $$x\lt 0.$$

Subcase 2a: If $$x\gt-1$$ then $$x+1\gt 0.$$ And thus $$\mathrm{x}^2+x+1\ge x+1\gt 0.$$

Subcase 2b: If $$x\le-1$$ then multiplying by x yields $$\mathrm{x}^2\ge-x$$ implying $$\mathrm{x}^2+x\ge 0$$. Thus $$\mathrm{x}^2+x+1\gt \mathrm{x}^2+x\ge 0.$$

Put $x=\dfrac{b}{a}$

$x^2+x+1=\dfrac{b^2}{a^2}+\dfrac{b}{a}+1$ $=\dfrac{b^2+ab+a^2}{a^2}=\dfrac{\dfrac{1}{2}(a+b)^2+\dfrac{b^2}{2}}{a^2}+\dfrac{1}{2}> 0$