Show that all real roots of the polynomial $P (x) = x^5 − 10x + 35$ are negative. I got this problem out of Andreescu's Putnam and Beyond. I solved it differently from the given solution and could not understand the solution. Can you explain what is happening in the last step of the solution?
Because P (x) has odd degree, it has a real zero r. If r > 0, then by the AM–GM
inequality
$P (r)$
$ = r^5 + 1 + 1 + 1 + 2^5 − 5 · 2 · r $
$≥ 0.$ (why?)
And the inequality is strict since $1 \neq 2$. Hence r < 0, as desired.

Also, here is my own (edit: incorrect) solution:
Let p be a positive root. Then $p^5-10p+35=0$
$10p > p^5 + 35 \implies 10 > p^4 + 35/p > 2 \sqrt{35p^3} > 10p \implies p < 1$
$10p > 35 \implies p > 3.5$
Contradiction!

That should say $P(r) = r^5+1+1+1+2^5-5\cdot 2 \cdot r$. 
The AM-GM inequality applied to $\{r^5,1,1,1,2^5\}$ gives us: 
$\dfrac{r^5+1+1+1+2^5}{5} \ge \sqrt[5]{r^5 \cdot 1 \cdot 1 \cdot 1 \cdot 2^5}$
$\dfrac{r^5+35}{5} \ge 2r$
$r^5+35 \ge 10r$
$r^5-10r+35 \ge 0$
Equality only holds if $r^5 = 1 = 1 = 1 = 2^5$, which gives $r = 1 = 2$, which is impossible. 
Thus, $r^5-10r+35 > 0$ for any number $r > 0$. Hence $x^5-10x+35$ cannot have a positive root. 
One error in your solution is that if $p$ is a root, then $10p = p^5+35$, so you can't say $10p > p^5+35$. Also, $2\sqrt{35p^3} > 10p$ is only true for $p > \dfrac{5}{7}$
