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
 A: 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 $a>0$ - 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.
A: Note that $a^5-a^3+a=3$ is equivalent to $0=a^5-a^3+a-3 =: f(a)$
You can easily see that the derivative $f'(a) = 5a^4-3a^2+1$ is strictly positive everywhere (I leave the proof to you with the hint $b := a^2$.) which implies that there is just one real root of $f$, let us call it $a_0$
Observe that $f(25/19)<0$. That means the root $a_0>25/19$. Therefore $a_0^6 > \left(\frac{25}{19}\right)^6 >5$.
A: 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 $3$: 
$$a^2 + 1 -2a \ge 0$$
$$(a-1)^2 \ge 0$$
A: The discriminant of $z^2-z+1$ is negative, hence $a(a^4-a^2+1)=3$ implies $a>0$.
Since $z^2-z+1$ attains its minimum at $z=\frac{1}{2}$, $f(z)=z^4-z^2+1$ attains its minimum, $\frac{3}{4}$, at $z=\frac{1}{\sqrt{2}}$. Moreover, $0<f(z)\leq 1$ if $z\in[0,1]$, hence $z\cdot f(z)=3$ implies $z>1$. 
$g(z)=z\cdot f(z)$ is an increasing function over $[1,+\infty)$ as the product of two positive increasing functions. Since $g(5^{1/6})<3$ by direct computation, in order to have $g(z)=3$
$$ z > 5^{1/6} $$
must hold. It may be useful to notice that:
$$ g(z) = \frac{z(z^6+1)}{z^2+1}, $$
hence if $z=5^{1/6}$ we have $g(z)=\frac{6z}{z^2+1}<3$ by the AM-GM inequality.
A: Define $f: \mathbf{R} \to \mathbf{R}: x \mapsto x^5-x^3+x-3$. Then $$
f^\prime(x)=5x^4-3x^2+1=5\left(x^2-\frac{3}{10}\right)^2+\left(1-\frac{9}{20}\right)>0.
$$
Therefore $f$ is strictly increasing, so it is sufficient to check that $f(5^{1/6})<0$.
