An algebra problem by A Former Brilliant Member

We will prove that $A> B$ . By cross multiplying, proving this inequality is equivalent to proving that $105\pi > 352 - 7\sqrt{10} .$

Since $f(x) = \dfrac{x^8(1-x)^8}{1+x^2}$ is strictly non-negative in the interval $[0,1]$ , the area bounded by this curve $f(x)$ , the $x$ -axis, and $0 \leq x \leq 1$ must be positive, thus $\displaystyle \int_0^1 \dfrac{x^8(1-x)^8}{1+x^2} \, dx = 4\pi - \frac{188684}{15015}$ must be a positive number. (The proof of the evaluation of the integral is left as an exercise for the readers)

Thus, $4\pi - \frac{188684}{15015} > 0$ , or $105\pi > \frac{47171}{143}$ . Now, we are left to prove that $\frac{47171}{143} > 352 - 7\sqrt{10}$ , which is equivalent to $1001 \sqrt{10} > 3165 .$ Squaring both sides shows that it's true. And we're done!

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How did I come up with that integral? See this .