How to prove this inequality given the equation Given $ a&gt;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
Given $a^5 - a^3 + a = 3$, we can factorise the left hand side to write $a(a^4-a^2+1)=3$. 

The second factor here is reminiscent of cyclotomic polynomials/finite geometric series - in any case, we can write it as $\frac{a^6+1}{a^2+1}$, and clear fractions to get to $a(a^6+1)=3(a^2+1)$.

Now, we spot the $a^6$ we're looking for on the left hand side, and rearrange to see that $a^6 = 3\frac{a^2+1}{a}-1=3(a+\frac{1}{a})-1$.

There are a number of ways of showing $a+\frac{1}{a}\geq2$ for $a&gt;0$ - the easiest is probably differentiating with respect to $a$ to find a minimum at $a=1$, but you can equally use the arithmetic-geometric inequality or a number of other methods. Substituting $a+\frac{1}{a}\geq2$ back into the previous inequality gets us to $a^6\geq3(2)-1=5$ and we're done.

We also could have, upon spotting the $a^6$, subtracted 5 and looked for a reason for it to be positive - in this case, we'd see that $a^6 - 5 = 3(a+\frac{1}{a})-6=3(a+\frac{1}{a}-2)$, and pushing through this a bit takes us to $\frac{3(a^2-2a+1)}{a}=\frac{3(a-1)^2}{a}$.

It's then clear that $a^6=5+\frac{3(a-1)^2}{a}$, which displays fairly immediately that the inequality holds.
Define $f: \mathbf{R} \to \mathbf{R}: x \mapsto x^5-x^3+x-3$. Then $$
f^\prime(x)=5x^4-3x^2+1=5\left(x^2-\frac{3}{10}\right)^2+\left(1-\frac{9}{20}\right)&gt;0.
$$
Therefore $f$ is strictly increasing, so it is sufficient to check that $f(5^{1/6})&lt;0$.