<s>[INST] Prove: $|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$ 
$$|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$$

I have tried: $$|a\sin x+b \cos x|\leq |a+b|\leq \sqrt{a^2+b^2}$$
enough to prove: $$|a+b|\leq \sqrt{a^2+b^2}$$
But I can find how to continue from here
 [/INST] If we square the inequality we get $$a^2\sin^2 x+b^2\cos^2 x+2ab\sin x\cos x\leq a^2+b^2\\2ab\sin x\cos x\leq a^2(1-\sin^2 x) +b^2(1-\cos^2 x) =a^2\cos^2x+b^2\sin ^2 x\\0\leq a^2\cos^2x-2ab\sin x \cos x+b^2\cos^2 x=(a\cos x-b\sin x) ^2$$
The last equation is always true, and we can square inequality because both sides are positive.
</s>