Algebraic proof that if $a>0$ then $1+a^9 \leq \frac{1}{a}+a^{10} $

Prove that if $a>0$ then $1+a^9 \leq \frac{1}{a}+a^{10} $.

Using the fact that $a \gt 0$, multiply by $a$ on both sides and get everything to one side we have; $a^{11}-a^{10}-a+1 \geq 0$. By factoring $(a^{10}-1)(a-1) \geq 0 $.

I am not sure how to proceed any further.

Solutions Collecting From Web of "Algebraic proof that if $a>0$ then $1+a^9 \leq \frac{1}{a}+a^{10} $"

If $a>0$ then from $a \le 1$ follows $a^{10}\le 1$ and from $a \ge 1$ follows $a^{10}\ge 1$ So either $(a-1)\ge0$ and $(a^{10}-1)\ge0$ or $(a-1)\le0$ and $(a^{10}-1)\le0$. In both cases we have $$(a-1)\cdot (a^{10}-1)\ge0$$

But for this special inequality we can argument in the following way:

With both methods you can easily deduce

if $a<-1$ then $(a^{10}-1)(a-1) \leq 0$


if $-1<a<0$ then $(a^{10}-1)(a-1) \geq 0$

We need to proof $1+a^9\le\frac{1}{a}+a^{10}$. You already proof that is sufficient and necessary that $(a-1)(a^{10}-1)\ge 0$.

This last inequality is true (when $a>0$) because:

And is trivial that $a^9+a^8+a^7+a^6+a^5+a^4+a^3+a^2+a+1> 0$, because $a>0$, and always $(a-1)^2\ge 0$, multiplying this two inequalities we are done.

One way is by evaluating cases. First look at $a\ge1$. We clearly have that $a-1$ and $a^{10}-1$ are greater than or equal to zero (If you want, let $a=1+\epsilon$, do a bit of work with the binomial theorem, and conclude). As such, multiplying the two gives a non-negative result and we are done.

For the second case we look at $0\le a<1$. It is clear that $a-1$ is negative, but the case $a^{10}-1$ is a bit trickier. The key thing to note here is that $a^n<a$ for any positive $n>1$ and $0<a<1$. Think of it this way: any number between $0$ and $1$ can be written as a fraction $\frac{1}{k}$ for some $k>1$, and so multiplying this number by itself makes the denominator larger and larger and thus makes the fraction smaller and smaller as a whole. As a result of this, $a^{10}<a$ here and thus $a^{10}-1 < a-1<0$. Because both terms are negative, we conclude that their product must be positive, and we are done.

For all $a>0$ we see that $\left(a^{10},1\right)$ and $\left(a,1\right)$ are the same ordered.

Thus, by Rearrangement $$a^{10}\cdot a+1\cdot1\geq a^{10}\cdot1+1\cdot a$$ and we are done!