Prove that $a+1 < a^2$ for all integers $a > 1$ I know it is true, but how could I prove it?
$$a^2-(a+1)>0$$
$$a^2 - a -1 >0$$
via a graphical solution $a^2-1-1>0$ when $a>$ approx $1.68$...thus given $a$ is an integer $>1$ the statement is true.
Can one do it without a graphical solution?
 A: Notice that $(x-1)^2\gt0$, since it's a square and $x\ne1$. Hence
$$x^2-2x+1\gt0$$
$$x^2\gt2x-1$$
And since $x\ge2$,
$$x^2\gt x+x-1\ge x+2-1$$
$$x^2\gt x+1$$
A: Since $x^2-x-1=(x-1)^2+(x-2)>0$ for $x\ge2$, $\;\;\;x^2>x+1$
A: No proof by induction has appeared, so here's one.
The statement is true for $a=2$, because $2+1<2^2$.
Suppose the statement is true for some integer $a\ge 2$; in other words, we assume that $a+1<a^2$; then
$$
(a+1)+1<a^2+1<a^2+2a+1=(a+1)^2.
$$
Fill in the details.
A: $x + 1 < x^2 \ \Longleftrightarrow \ x^2 - x - 1 > 0 \ \Longleftrightarrow \ x^2 - x +1/4 - 5/4 > 0 \ \Longleftrightarrow$
$$(x -1/2)^2 > 5/4$$
which is implied by
$$x - 1/4 > \sqrt{5}/2$$
This last relation is true as $x \geq 2 > \sqrt{5}/2 + 1/4$
A: We have
$$
x^2>x+1\\
x^2-x-1>0\\
x^2-2x+1+x-2>0\\
(x-1)^2+x-2>0
$$
where the last one is obviously true if $x\geq2$.
A: The real roots of $x^2 - x -1=0$ are $\dfrac{1\pm \sqrt{5}}{2}$. So, $x^2 > x+1$ forall real numbers greater than $\dfrac{1+ \sqrt{5}}{2} $. But $ \dfrac{1+ \sqrt{5}}{2} <2 $  so the statement holds for integers greater than $1$.
A: If $x\ge2$, then
$${1\over x}+{1\over x^2}\lt{1\over x}+{1\over x}\le{1\over 2}+{1\over2}=1$$
Multiplying both sides by the positive quantity $x^2$ gives the desired inequality, $x+1\lt x^2$.
A: $$ a+1 < a+a = 2a \le aa = a^2 $$
A: $$f(x)=x^2,\quad g(x)=x+1$$
$$f(2)>g(2)$$
$$f'(x)=2x > g'(x)=1\quad(x\geq2)$$
$$\therefore x^2=f(x)>g(x)=x+1\quad(x\geq2)$$
A: If $x=2$, then $x^2=4 > 3=x+1$.  And if $x^2>x+1$ and $x > 1$, then
$$
(x+1)^2=x^2+2x+1 >(x+1)+2x+1 > (x+1)+1.
$$
This completes the proof for all integers $x > 1$ by induction.
A: The inequality you want to show is equivalent to
$$
\begin{align*}
a^2-1 &> a\\
(a-1)(a+1) &> a
\end{align*}
$$
If $a\ge2$, then $a-1\ge1$ and you get
$$(a-1)(a+1)\ge a+1 >a.$$ 
