<s>[INST] 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!
 [/INST] To see what can be done without calculus, we can use a little bit from the "theory of equations" to show that $ \ x^5 \ - \ 10x \ - \ 35 \ $ has either five, three, or one real roots (complex conjugate zeroes) and that either two or none of those are positive and just one is negative (Descartes' Law of Signs).
If we just work with $ \ x^5 \ - \ 10x  \ = \ x \ (x^2 \ - \ 10^{1/2} ) \ (x^2 \ + \ 10^{1/2} ) $ , we have an even function with three real zeroes.  From what we know about the form of such a curve ("negative tail" goes to negative infinity, "positive tail" to positive infinity), between $ \ x \ = \ -(10^{1/4}) \ $ and $ \ x \ = \ 0 \ $ lies the local maximum for this function and the local minimum lies between  $ \ x \ = \ 0 \ $ and $ \ x \ = \ +(10^{1/4}) \ $  This latter value of $ \ x \ $ is about $ \ 3.2^{1/2} \ \approx \ 1.8 \ $ (since $ \ \sqrt{10} \ \approx \ 3.2 \ $  and $ \ \sqrt{3.24} \ = \ 1.8 \ $ ) .  
We can now estimate the minimal value of the function, using values of $ \ x \ $ between $ \ 1 \ $ and $ \ 1.8 \ $ , by noting that
$$ P \ (1) \ = \ 1^5 \ - \ 10 \ = \ -9 \ \ , $$
$$ P \ ( \ \frac{3}{2} \ ) \ = \ \left( \frac{3}{2} \right)^5 \ - \ 10 \ \left( \frac{3}{2}  \right) \ = \ \left( \frac{243}{32}  \right) \ - \ 15 \ \sim \ 8 \ - \ 15 \ \sim \ -7  \ \ , $$
$$ P \ ( \ \frac{4}{3} \ ) \ = \ \left( \frac{4}{3} \right)^5 \ - \ 10 \ \left( \frac{4}{3}  \right) \ = \ \left( \frac{1024}{243}  \right) \ - \ \frac{40}{3} \ \sim \ 4 \ - \ 13 \ \sim \ -9  \ \ , $$
where small integer ratios are chosen to keep the calculations simple.  It is clear that the local minimal value is probably not significantly different from $ \ -10 \ $ (in fact, it's about $ \ -9.51 \ $ ) .  
Shifting the curve for $ \ x^5 \ - \ 10x  \ $ "vertically" by 35 units then guarantees that $ \ P \ (0) \ $ and the local minimum of $ \ P(x) \ $ are positive.  So  only the single negative $ \ x-$ intercept remains. (The other four zeroes are in complex conjugate pairs.)
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