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.  

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:
$(a-1)(a^{10}-1)=(a-1)^2(a^9+a^8+a^7+a^6+a^5+a^4+a^3+a^2+a+1)$
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.
