When is $a^2 + b^2 + c^2 > ac\cos{x} + bc\sin{x}$? I was hoping someone could help me with finding the values of $a$, $b$, $c$ and $x$ for which the following inequation is true: $$a^2 + b^2 + c^2 > ac\cos{x} + bc\sin{x}$$
where $a$, $b$ and $c$ are real numbers and $0 \le x < 2\pi.$ 
I don't really know how to start and I think this problem is more complicated than it looks.
Thank you very much for reading.

$f(c) = c^2 - (a\cos x + b \sin x)c + a^2+b^2$. So $\triangle = (a\cos x + b\sin x)^2 - 4(a^2+b^2)\le (a^2+b^2)(\cos^2 x + \sin^2 x) - 4(a^2+b^2)\le - 3(a^2+b^2) \le 0$. Thus $f(c) \ge 0$ for all $c$. This means your inequality is true for all $a,b,c,x$. The minor cases are $a = 0$, then $b^2+c^2 \ge 2|bc| \ge |bc| \ge |bc\sin x| \ge bc\sin x$ with equality occurs when $b =c = 0$. Note that if $\triangle = 0 \implies a = b = 0$, and the inequality is true accept for $c = 0$ when equality occurs.
