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I am preparing a paper on Huygens' approximations to $\pi$ and I have to prove the following inequality: if $0 \le x\le \pi/2$ then $$ \pi\ge \frac{\pi}{x}\sin(x)\bigl(20+51\cos(x)-2\cos^2(x)+6\cos^3(x)\bigr)/\bigl(21+18\cos(x)+36\cos^2(x)\bigr). $$ The usual method of showing that the left hand side equals $\pi$ for $x=0$ and then trying to prove that it decreases for $x>0$ by proving that the derivative is negative is difficult since to apply the derivative is quite complex.

I would appreciate any help.

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  • $\begingroup$ The taylor expansion at x=0 is given by pi-pi x^6/175-O(x^8) $\endgroup$
    – Mark B Villarino
    Commented Feb 14 at 19:10
  • $\begingroup$ Maybe you missed a division sign between two brackets? $\endgroup$
    – Fedor Petrov
    Commented Feb 14 at 20:27
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    $\begingroup$ Do you have a response to the answer below? $\endgroup$
    – Iosif Pinelis
    Commented Feb 18 at 2:37

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Using the substitutions $$\tan\frac x2=t,\quad x=2\arctan t,\quad \cos x=\frac{1-t^2}{1+t^2},\quad \sin x=\frac{2t}{1+t^2}, $$ rewrite the inequality in question as $$d(t):=\frac{2 t \left(39 t^6-29 t^4-95 t^2-75\right)}{3 \left(t^2+1\right)^2 \left(13 t^4-10 t^2+25\right)}+2 \arctan t\ge0$$ for $t\in[0,1]$.

It remains to note that $d(0)=0$ and $$d'(t)=\frac{128 t^6 \left(13 t^4+40 t^2+75\right)}{\left(t^2+1\right)^3 \left(13 t^4-10 t^2+25\right)^2}\ge0$$ for all real $t$.

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