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

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.
