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:
$x^2+x+1=(x+\frac12)^2+\frac34$, which is always greater than $0$.
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 $0$.
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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.
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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.)
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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 $y_{min}= -\frac{1}{4}+1 = \frac{3}{4}>0$
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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.
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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.
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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
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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}
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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 $y>0$:
$y^2-2y + 1 +y =(y-1)^2 +y >0$
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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 $$x^2+x+1=x(x+1)+1>0+1>0.$$
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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.
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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 $x^2+x+1=1\gt 0$
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Case 1: $x\ge 0 \Rightarrow x^2+x+1\ge 0^2+0+1>0$.
Case 2: $x<0 \Rightarrow x^2+x+1>0 \iff \frac{x^2+x+1}{x}<0 \iff x+\frac 1x+1\overbrace{\le}^{AM-GM} -2+1<0$.
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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.
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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 $x=-1/2.$
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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 $f(x)\gt0\;\;\forall x\in\mathbb{R}$.
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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 $x$.
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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.
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Put $x=\dfrac{b}{a}$
$x^2+x+1=\dfrac{b^2}{a^2}+\dfrac{b}{a}+1$ $=\dfrac{b^2+ab+a^2}{a^2}=\dfrac{\dfrac{1}{2}(a+b)^2+\dfrac{b^2}{2}}{a^2}+\dfrac{1}{2}> 0$
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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 $\mathrm{x}^2+x+1\gt \mathrm{x}^2+x\ge 0.$
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 $>0$).