Weird Expression

We have

$a^2+b^2+\dfrac {1}{a^2} + \dfrac {b}{a} = \left (b + \dfrac {1}{2a} \right )^2 + a^2 + \dfrac {3}{4a^2}$

Since $\left ( b + \frac {1}{2a} \right )^2 \geq 0$ , we have

$a^2+b^2+\dfrac {1}{a^2} + \dfrac {b}{a} \geq a^2 + \dfrac {3}{4a^2}$

Using AM-GM, we have

$a^2 + \frac {3}{4a^2} \geq 2 \sqrt{a^2 \dfrac {3}{4a^2}} = \sqrt{3}$

Thus, the minimum value of the expression is $\sqrt{3}=1.732$ . This equality occurs if $b = - \frac{1}{2a}$ and $a^2=\frac {3}{4a^2}$ , so $a=\pm \sqrt[4]{\frac{3}{4}}$ and $b=\mp \frac {1}{2}\sqrt[4]{\frac{4}{3}}$ .

$x=a^{2}+b^{2}+\dfrac{1}{a^2}+\dfrac{b}{a}=b^{2}+\dfrac{1}{a}b+a^{2}+\dfrac{1}{a^2}$

The minimum value of $x$ happen when $b=-\dfrac{\frac{1}{a}}{2 \times{1}}=-\dfrac{1}{2a}$

$\begin{aligned} b=-\dfrac{1}{2a} \implies x&=(-\dfrac{1}{2a})^{2}+\dfrac{1}{a}.(-\dfrac{1}{2a})+a^{2}+\dfrac{1}{a^2} \\ x&=\dfrac{1}{4a^2}-\dfrac{1}{2a^2}+a^{2}+\dfrac{1}{a^2} \\ x&=a^{2}+\dfrac{3}{4a^2} \end{aligned}$

By using $AM \geq GM$ , we have

$\begin{aligned} \dfrac{a^2 +\frac {3}{4a^2}}{2} &\geq \sqrt{a^2 \dfrac {3}{4a^2}} \\ a^2 + \frac {3}{4a^2} &\geq \sqrt{3} \\ x &\geq \sqrt{3} \\ x &\geq 1.732 \end{aligned}$

Then, the minimum value of $x$ is $1.732$

We can solve this question without using the AM-GM inequality:

$a^2 + b^2 + \frac {1}{a^2} + \frac {b}{a} = (a^2 - \sqrt {3} + \frac {3}{4a^2}) + (b^2 + \frac {b}{a} + \frac {1}{4a^2}) + \sqrt {3} =$

$= (a - \frac { \sqrt {3} }{2a})^2 + (b + \frac {1}{2a})^2 + \sqrt {3}$

Now, what we have is a sum of two squares and a constant. A square is minimal (on its own), when its value is 0.

In this case, we can make both squares 0 at the same time:

$a - \frac { \sqrt{3} }{2a} = 0$

$a^2 = \frac { \sqrt{3} }{2}$

$a = ± \sqrt [4]{ \frac {3}{4}}$

$b + \frac {1}{2a} = 0$

$b = - \frac {1}{2a}$

$b = ± \frac {1}{2} \sqrt [4]{ \frac {4}{3}}$

Hence, the minimum value of our weird expression is our constant:

$\boxed { \sqrt{3} = 1.732 \text { ( 3 d. p. ) } }$