If the minimum value of $f\left(x\right)=\frac{5x^2}{2}+\frac{\alpha }{x^5},x>0$, is $14$, then the value of $\alpha$ is equal to

From weighted $\mathrm{A}.\mathrm{M}.-\mathrm{G}.\mathrm{M}.$ inequality, we know

$\frac{5\left(\frac{x^2}{2}\right)+2\left(\frac{\alpha }{2x^5}\right)}{5+2}\ge {\left({\left(\frac{x^2}{2}\right)}^5·{\left(\frac{\alpha }{2x^5}\right)}^2\right)}^{\frac{1}{7}}$

$\Rightarrow \frac{5x^2}{2}+\frac{\alpha }{x^5}\ge \frac{7}{2}{\left(\alpha \right)}^{\frac{2}{7}}$

Given that the least value of $\frac{5x^2}{2}+\frac{\alpha }{x^5} \text{is} 14$

i.e. $\frac{7}{2}{\left(\alpha \right)}^{\frac{2}{7}}=14\Rightarrow {\left(\alpha \right)}^{\frac{1}{7}}=2\Rightarrow \alpha =128$