<s>[INST] Prove, that $\sin x- a^3\cos x\leq \frac 1 3 \sqrt{1+a^6}$ Let a and $x$ be natural numbers with the property that $\sin x\leq a\cos x$.
Prove that $\sin x- a^3  \cos x\leq \frac 1 3 \sqrt{1+a^6}$.
Again, I'm looking for a second solution. I don't know how to use LaTex and my solution is hard to write here. It would be very nice to see a different solution from mine.
 [/INST] Consider the problem 
$$\max_{\mathbf{y}\in \mathbb{R}^2} \mathbf{y} \cdot \mathbf{c}$$ subject to $\mathbf{b}\cdot \mathbf{y} \le 0$ and $\|\mathbf{y}\|=1$, where $\mathbf{b}=(-a,1)$, $\mathbf{c}=(-a^3,1)$. Since $\mathbf{b} \cdot \mathbf{c} \ge 0$ it's not too hard to see   that the optimal argument is given by $\mathbf{y}^*=(\cos x,\sin x) = (-1,-a)/\sqrt{1+a^2}$. (Drawing a picture of the feasible region helps.) So then we want to check that:
$$\mathbf{y}^*\cdot \mathbf{c} = \frac{a^3-a}{\sqrt{1+a^2}} \le \frac{1}{3}\sqrt{1+a^6}$$
Since $a \ge 1$ this is equivalent to:
$$9 a^2(a^2-1)^2 \le (1+a^2)(1+a^6).$$
Which then by collecting terms and factoring, is equivalent to:
$$a^8-8a^6+18a^4-8a^2+1=(a^4-4a^2+1)^2\ge 0.$$
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