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

Question

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 ,$t=\sqrt{2}$.

Answer 1

The problem is that for AM-GM equality holds only when

$$\frac{1}{t^4}=\frac{3}{t^2}= \frac{t^2}{3} \iff1=3t^2=\frac{t^6}3$$

which is not possible, therefore the estimation is correct but it is not helpful to find the minimum.

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As an alternative to evaluate the minimum we have that by $t=\tan x$

$$a=\frac{{(1+t^2)}^3}{t^4}=\frac{1}{\sin^4 x\cos^2 x}=\frac{1}{\cos^2 x(1-\cos^2 x)^2}$$

and the problem reduces to study the maximum for $u^2(1-u)$ with $0<u<1$ with

$$u^2(1-u)=\frac4{27}-\left(u-\frac 2 3\right)^2-\left(u-\frac 2 3\right)^3\le \frac 4{27}$$

Answer 2

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 $t=-\sqrt 2$.

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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 $1=\frac{t^2}2$.)

Answer 3

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 $t = \pm \sqrt{2}.$