Prove this inequality $ \sqrt{5} > \frac {13 + 4\pi}{24 - 4\pi} $

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$$ \sqrt{5} > \frac {13 + 4\pi}{24 - 4\pi} $$

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This is the same as proving that $$\frac{1}{16}(133-37\sqrt{5})>\pi$$ and it follows from the fact that the continued fraction of the LHS is: $$ [3;7,15,1,660,\ldots] $$ while the continued fraction of $\pi$ is: $$ [3;7,15,1,292,\ldots].$$