Minimum of a function without calculus. $a=\frac{{(1+t^2)}^3}{t^4}$ 
Find the  the minimum value of $a$. .$$a=\frac{{(1+t^2)}^3}{t^4}$$.

Instead of calculus i tried using the AM-GM inequality.,as follows:we have $$3+\frac{1}{t^4}+\frac{3}{t^2}+3\left( \frac{t^2}{3}\right)\ge 3+{\left(\frac{1}{9}\right)}^{1/5}$$ which is not the minimum i got by using calculus.What mistake am i making using this inequality? And also it would be very helpul if anyone could arrive at the minima using basic inequalities.(withoutv calculus)

using calculus  minima occured at ,$t=\sqrt{2}$.


Using the AM-GM inequality, we have
$$1+t^2 = 1+\frac{t^2}{2}+\frac{t^2}{2} \geqslant 3\sqrt[3]{1 \cdot \frac{t^2}{2} \cdot \frac{t^2}{2}} = 3\sqrt[3]{\frac{t^4}{4}},$$
therefore
$$(1+t^2)^3 \geqslant \frac{27}{4}t^4,$$
or
$$\frac{(1+t^2)^3}{t^4}\geqslant \frac{27}{4}.$$
Equality occur when $\frac{t^2}{2} = 1,$ or $t = \pm \sqrt{2}.$
