Prove that $a^2+\dfrac{1}{a^a-a+1}\ge a+1$, for any number $a$

Question

Prove that $a^2+\dfrac{1}{a^a-a+1}\ge a+1$, for any number $a$

I think I can solve the problem by putting all the variables on one side and then $0$ on the other side, then factoring the side with the variables to become a square. This proves the inequality because anything squares is positive. Can I solve this problem using algebra? Obviously you can solve it using induction.

Answer 1

This is probably not the most complete answer but it (somehow) proves the claim.

First, note that for $a\geq 1$, we have $$a^a-a+1 \geq a^a \geq a\geq 1$$ so that $\frac{1}{a^a-a+1}\geq 0$.

Now, $a^2-a+1$ is a convex parabola and $a^2-a-1=0$ for $a=\phi$ where $\phi$ is the golden ratio $\phi = \frac{1+\sqrt{5}}{2}$. It follows that $$ a^2+\frac{1}{a^a-a+1}-(a+1)\geq a^2-a-1 \geq 0 \qquad \forall a \in [\phi,\infty).$$

I propose to conclude with a proof by drawing, indeed for $a\in[1,\phi)$ the graph of the continuous function $f(a)=a^2+\frac{1}{a^a-a+1}-a-1$ is given by:

and therefore your inequality is true for every $a\geq 1$.

Answer 2

Let $f:[1,\infty) \to \mathbb{R}$ be defined by

$$f(a) = a^2 - a - 1 + \frac{1}{a^a - a + 1}$$

The goal is to show $f(a) \ge 0$ for all $a \in [1,\infty)$.

First suppose $a \ge 2$. Then

\begin{align*} f(a) &= a^2 - a - 1 + \frac{1}{a^a - a + 1}\\[6pt] &> a^2 - a - 1\\[6pt] &\ge 2^2 - 2 - 1 = 1\\[6pt] &> 0\\ \end{align*}

Next suppose $1 \le a < 2$. Then

\begin{align*} f(a) &= a^2 - a - 1 + \frac{1}{a^a - a + 1}\\[6pt] &\ge a^2 - a - 1 + \frac{1}{a^2 - a + 1}\\[6pt] &= \left(a^2 - a + 1 + \frac{1}{a^2 - a + 1}\right) - 2\\[6pt] &\ge 2 - 2\qquad\text{[by AM-GM]}\\[6pt] &=0\\ \end{align*}

Therefore $f(a) \ge 0$ for all $a \in [1,\infty)$, as required.