Succinct proof that $\frac\pi4+\frac\pi6+\log2\gt2$

Question

In answering Average angle between two randomly chosen vectors in a unit square, I noticed that the average angle formed by two vectors uniformly picked in the unit square, $\frac\pi4+\log2-1\approx0.4785$, and the average angle formed by two vectors uniformly picked in the first quadrant of the unit disk, $\frac\pi6\approx0.5236$, add up to just a bit more than $1$ (about $1.0021$), that is, that

$$ \frac\pi4+\frac\pi6+\log2\gtrsim2\;. $$

I thought it would be interesting to try to prove this with as little numerics as possible. Of course you can use sufficiently good rational approximations for $\pi$ and $\log2$, like I did to answer Prove that $e^\pi+\frac{1}{\pi} < \pi^e+1$, but I’d prefer a proof that shows directly that $\frac\pi4+\frac\pi6+\log2-2$ is a positive quantity, e.g. the integral over a positive function, as in Is there an integral that proves $\pi > 333/106$?.

Answer 1

I originally just wanted to ask the question and didn’t expect to find an answer myself, but playing around with the exponents in

$$ \int_0^1\frac{x^4(1-x)^4}{x^2+1}\mathrm dx=\frac{22}7-\pi $$

I happened upon

$$ \int_0^1\frac{x^5(1-x)^4}{\left(x^2+1\right)^2}\mathrm dx=\pi+6\log2-\frac{73}{10}\;, $$

which is exactly the sort of thing we need. Together with

$$ \int_0^1\frac{x^2(1-x)^4}{x^2+1}\mathrm dx=\pi-\frac{47}{15} $$

(see Integrals for semiconvergent approximations to $\pi$), this yields

\begin{eqnarray} \frac16\int_0^1\frac{x^5(1-x)^4}{\left(x^2+1\right)^2}+\frac14\int_0^1\frac{x^2(1-x)^4}{x^2+1}\mathrm dx &=& \frac1{12}\int_0^1\frac{\left(2x^3+3x^2+3\right)x^2(1-x)^4}{\left(x^2+1\right)^2}\mathrm dx \\ &=& \frac\pi4+\frac\pi6+\log2-2\;. \end{eqnarray}