How to prove this innocent inequality?

If I remember it right, I saw this inequality long ago, when I was a schoolboy. Then I didn’t solve it. But now it seems the following for me.

Of course, the geometric approach is the first solution idea, :-) but it should be developed. Without loss of generality, it suffices to consider Quadrant I of the plane (the disposition in the Quadrant III is symmetric).

We try to find the parallel lines separating the curves. The experimental evidence suggests to try $2a+3b=x$ and $2d+3c=y$ (see the picture).

The first line. In the coordinates $(a/2,b)$ the line is a tangent to the unit circle. It these coordinates the line has the equation $4(a/2)+3b=x$. Therefore it is tangent to the circle at the point $(4,3)/\sqrt{4^2+3^2}=(4/5,3/5)$. Then $x=(4\cdot (4/5)+3\cdot (3/5))=5$.

The second line. $2d+3b\ge 2\sqrt{2d\cdot 3b}=4\sqrt 6=y$.

The squared distance between the lines is $\frac {(x-y)^2}{2^2+3^2}=1.77\dots>1.6$.

If $a^2+4b^2=4$ , we have an ellipse with width $2$ and height $1$. Now, the set of points $cd=4$ describes a hyperbola which does not cross the ellipse. You need to prove that: $$(a-d)^2 + (b-c)^2 \geq 1.6$$ This is equation for a circle centered around the point $(c,d)$ which lies on the hyperbola. In other words, your problem is to prove that the hyperbola is always at least $1.6$ units away from the ellipse.

Therefore, it is enough to find the minimal distance between the functions: $$f(x) = \frac{4}{x},\ \ g(x) = \sqrt{\frac{4 - x^2}{4}}$$ And to show that this is larger than $1.6$, i.e. find the minimum of:

$$\min_{x_1,x_2} \sqrt{\left( \frac{4}{x_1} - \sqrt{\frac{4 - x_2^2}{4}} \right)^2 -(x_1-x_2)^2}$$ In the range $x_2\in[0,2]$, $x_1>0$.