How to prove this inequality given the equation Given $ a>0 $ and $ a^5-a^3+a=3$, how can I prove the inequality: $ a^6 \ge 5 $ ?
I have tried factorizing the equation, solving for $ x, x^2, x^3 $ and then adding equations made together. Last but not least I tried, to solve for $x$ and then raise it to the power of $ x^6 $, but none of these seem to work

I actually found a simpler solution to my problem:

$$a^5 -a^3 + a = 3 $$
Multiply by $a$: 
$$a^6 - a^4 + a^2 = 3a $$

$$a^6 = a^4 - a^2 + 3a \ge 5 $$
Multiply by $a$:
$$ a^5 - a^3 + 3a^2 \ge 5a$$

Changing the first equation: $$a^5 - a^3 = 3-a$$
Then: $$3-a + 3a^2 \ge 5a$$
$$3a^2 + 3 - 6a \ge 0$$
Divide by $3$: 
$$a^2 + 1 -2a \ge 0$$
$$(a-1)^2 \ge 0$$
