Nice Question:

let $x\in [0,2\pi]$, show that:


I know this follow famous problem(1995 Russia Mathematical olympiad)


This problem solution can see :http://iask.games.sina.com.cn/b/19776980.html and everywhere have solution in china BBS

I post this problem solution

case1: if $x\in[\pi,2\pi]$,then $$\cos{\cos{\cos{\cos{x}}}}>0,\sin{\sin{\sin{\sin{x}}}}\le 0$$ so $$\cos{\cos{\cos{\cos{x}}}}>\sin{\sin{\sin{\sin{x}}}}$$

case2: if $x\in[0,\dfrac{\pi}{2}]$,then we have $$\cos{x}+\sin{x}\le\sqrt{2}<\dfrac{\pi}{2}\Longrightarrow 0\le \cos{x}<\dfrac{\pi}{2}-\sin{x}$$ so $$\cos{\cos{x}}>\cos{\left(\dfrac{\pi}{2}-\sin{x}\right)}=\sin{\sin{x}}$$ $$\sin{\cos{x}}<\sin{\left(\dfrac{\pi}{2}-\sin{x}\right)}=\cos{\sin{x}}$$ then $$\cos{\cos{\cos{x}}}<\cos{\sin{\sin{x}}}$$ so $$\cos{\cos{\cos{x}}}+\sin{\sin{\sin{x}}}<\cos{\sin{\sin{x}}}+\sin{\sin{\sin{x}}}<\dfrac{\pi}{2}$$ so $$\cos{\cos{\cos{x}}}<\dfrac{\pi}{2}-\sin{\sin{\sin{x}}}$$ then $$\cos{\cos{\cos{\cos{x}}}}>\cos{\left(\dfrac{\pi}{2}-\sin{\sin{\sin{x}}}\right)}=\sin{\sin{\sin{\sin{x}}}}$$ case3: if $x\in (\dfrac{\pi}{2},\pi)$,then let

$y=x-\dfrac{\pi}{2}$,so $$\cos{\cos{\cos{\sin{y}}}}>\sin{\sin{\cos{\sin{y}}}}$$ and since $f(t)=\sin{\sin{t}}$ is increasing,then $$f(\cos{\sin{y}})>f(\sin{\cos{y}})\Longrightarrow \sin{\sin{\cos{\sin{y}}}}>\sin{\sin{\sin{\cos{y}}}}$$ so $$\cos{\cos{\cos{\sin{y}}}}>\sin{\sin{\sin{\cos{y}}}}$$ so $$\cos{\cos{\cos{\cos{x}}}}>\sin{\sin{\sin{\sin{x}}}}$$

But I found this $\dfrac{4}{5}$ maybe is strong,

so if $x\in[\pi,2\pi]$,then we have $$\dfrac{4}{5}\cos{\cos{\cos{\cos{x}}}}\ge 0>\sin{\sin{\sin{\sin{x}}}}$$

But for the case $x\in [0,\pi]$, I can't prove this $$4\cos{\cos{\cos{\cos{x}}}}\ge 5\sin{\sin{\sin{\sin{x}}}}$$

Thank you very much!

  • $\begingroup$ They don't seem to cross each other, but get very close (about $0.004$ apart) at around $x=1.159$ and $x=1.983$… so it might be tough to separate them analytically. $\endgroup$
    – mjqxxxx
    Dec 2, 2013 at 15:40
  • $\begingroup$ You can replace $\frac45$ by $.7950698563775$, it seems strict inequality still holds. $\endgroup$
    – Macavity
    Dec 2, 2013 at 16:16
  • $\begingroup$ The Maple command $$Optimization:-Minimize((4/5)*cos(cos(cos(cos(x))))-sin(sin(sin(sin(x)))), x = 0 .. 2*Pi)$$ outputs $$ [0.00405634334541515874,[x= 1.15850191101073886]] .$$ This is a modern approach to such type problems. $\endgroup$
    – user64494
    Dec 2, 2013 at 16:19
  • 1
    $\begingroup$ It is sufficient to prove the inequality $$ s(x)=\cos^{(4)}(\pi/2)-\sin^{(4)}(x+\pi/2)\geq \frac{2}{5}\sin(1)\sin(\cos 1)\, x^2\geq \cos^{(4)}(\pi/2)-\cos^{(4)}(x+\pi/2)=c(x) $$ over the interval $I=[-\pi/2,\pi/2]$. Since the middle term is the truncated Taylor series of the right term, to prove the second inequality it is sufficient to show that $c(\sqrt{x})$ is a concave function over $[0,\pi^2/4]$. The left inequality is harder, since $s(\sqrt{x})$ is convex only between $0$ and $1.9221$, and the difference between the first term and the second is just $0.001886$ in the "tightest" point. $\endgroup$ Dec 5, 2013 at 14:02
  • 1
    $\begingroup$ In principle one can prove/verify this inequality by using the fact that, if we set $f(x)=\frac{4}{5}\cos\cos\cos\cos x-\sin\sin\sin\sin x$, we clearly have $|f'(x)|<2$, so that $f$ is Lipschitzian with Lipschitz constant 2. So it sufficies to prove that $f(x_n)>\frac{4}{1000}$ for $x_n=\frac{4n}{1000}$ ($n=0,..,785$) and we have that for any $x\in [0,\pi]$ there exists $n$ such that $|x-x_n|\le\frac{2}{1000}$ and we are done. I don't think there are nice methods which avoid a great deal of computation since (as it has been pointed out) this inequality is very sharp. $\endgroup$
    – Mizar
    Dec 8, 2013 at 12:53

2 Answers 2


We still start from the original Russian Olympiad Problem: $\cos \cos \cos \cos x> \sin \sin \sin \sin x$. It could have another numerical proof simply by doing in a calculator:

We have $-1\leq \cos x \leq 1, \text{that is }\cos 1\leq \cos \cos x\leq 1, \text{that is }\cos 1 \leq \cos \cos \cos x \leq \cos \cos 1$.

Finally, we have,

$$ ~0.6542 \simeq \cos \cos \cos 1\leq \cos \cos \cos \cos x \leq \cos \cos 1. $$

Similarly, we have

$$ -\sin \sin \sin \sin 1 \leq \sin \sin \sin \sin x \leq \sin \sin \sin 1 \simeq0.6784... $$

If the equation has the solution, that is

$$ \sin x\geq \sin^{-1} \sin^{-1}\sin ^{-1} \cos \cos \cos 1 \simeq 0.6086... $$

Thus, we have $$ |\cos x|\leq 0.7835... $$

Therefore, $\cos \cos \geq 0.7013 \to \cos \cos \cos x \leq 0.7639 \to \cos \cos \cos \cos \cos x \geq 0.7221...$

Thus, it is not possible to have $\sin \sin \sin \sin \sin \leq 0.6784.$ The inequality holds.

So, the $\frac{4}{5}$ is still not a strong constant, and inequality proof as similar.


Incomplete, but too long for a comment:

$\displaystyle f(x)\leqslant a\cdot g(x)\iff h(x)=\frac{f(x)}{g(x)}\leqslant a\iff$ We have to compute all solutions to $h'(x)=0$ , calculate the value of $h(x)$ in those points, as well as in $0$ and $2\pi$, which are the extremities of the interval, and verify $h(x_{_\text{k}})\leqslant a$, for all these $x_{_\text{k}}$ . Due to the function's parity and periodicity (cosine at the denominator, and an even number of sines in the numerator), the interval may just as well be restricted to $\Big[0,\frac\pi2\Big]$, since it can quite easily be shown that $h(x)=h(\pi-x)=-h(\pi+x)=$ $=h(2\pi+x)$. Either way, all its extrema are of the form $h(k\pi\pm x_0)$ and $h\left((2k+1)\frac\pi2\right)$, where $x_0\simeq1.1631454$ is the only solution to the transcendental equation $h'(x)=0$ on the open interval $\left(0,\frac\pi2\right)$. Indeed, $h\left((2k+1)\frac\pi2\right)\simeq0.791<h(k\pi\pm x_0)\simeq0.795<\displaystyle\frac45$. Now, that $h'(x)$ has roots in odd multiples of $\frac\pi2$ is trivial to show, but proving the uniqueness of $x_0$ on $\left(0,\frac\pi2\right)$ is anything but.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.