<s>[INST] 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 ,$t=\sqrt{2}$.

 [/INST] A very short proof that the minimal value is $\frac{27}4$ can be given by noting that $$(1+t^2)^3-\frac{27}4 t^4=\frac{(\sqrt 2 - t)^2 (\sqrt 2 + t)^2 (1 + 4 t^2)}4\geq0$$
with equality if and only if $t=\sqrt 2$ or $t=-\sqrt 2$.

EDIT: Systematically we can proceed as follows: For any $a>0$, we have by weighted AM-GM, $$1+t^2=1+ a\frac{t^2}a\geq(1+a)\left(\frac{t^2}a\right)^\frac{a}{1+a}.$$
We want the power of $t$ on the right-hand side to be $\frac43$, so that if we raise it to the third power, it is of the same power as $t^4$. For this, we have to have $\frac{a}{1+a}=\frac23$ which is equivalent to $a=2$. If we use the inequality for $a=2$, we get $$1+t^2\geq\frac{3}{2^\frac23} t^\frac43.$$
It follows that $$(1+t^2)^3\geq \frac{3^3}{2^2} t^4=\frac{27}4 t^4.$$
(Equality iff $1=\frac{t^2}2$.)
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