Inequality Challenge: Prove $x$ for $x>0$

[anemone]

Prove $x+x^9+x^{25}<1+x^4+x^{16}+x^{36}$ for $x>0$.

 

[topsquark]

Let's see if my head is fully back in the game.

Spoiler

Let's look at \(\displaystyle f(x) = x^{36} - x^{25} + x^{16} - x^9 + x^4 - x + 1\)

For x = 0, f(0) = 1.

For 0 < x < 1:
\(\displaystyle x^{36} - (x^{25} - x^{16} ) - (x^9 - x^4) - (x - 1) > 0\) because each term inside the parentheses are negative.

For 1 < x
\(\displaystyle (x^{36} - x^{25}) + (x^{16} - x^9) + (x^4 - x) + 1 > 0\) because every term inside the parentheses are positive.

Therefore f(x) has no real zeros on 0 < x.

\(\displaystyle x^{36} - x^{25} + x^{16} - x^9 + x^4 - x + 1 > 0 \implies x^{36} + x^{16} + x^4 + 1 > x^{25} + x^9 + x\) on 0 < x.

-Dan