TITLE: Smallest positive integral value of $a$ such that ${\sin}^2 x+a\cos x+{a}^2>1+\cos x$ holds for all real $x$
QUESTION [7 upvotes]: If the inequality $${\sin}^2 x+a\cos x+{a}^2>1+\cos x$$ holds for all $x \in \Bbb R$ then what's the smallest positive integral value of $a$?

Here's my approach to the problem $$\cos^2 x+(1-a)\cos x-a^2<0$$
Let us consider this as a quadratic form respect to $a$.
Applying the quadratic formula $a=\frac{-\cos x\pm\sqrt{5\cos^2 x+4\cos x}}2 $
and substituting $\cos x$ with $1$ and $-1$
we get 3 values of where the graph should touch the x axis $-2,0,1$
How should I proceed now?

REPLY [3 votes]: We need that the inequality $\cos^2x+(1-a)\cos{x}-a^2<0$
will be true for all real $x$.
Let $\cos{x}=t$.
Thus, we need to find a smallest natural $a$ for which the inequality 
$$t^2+(1-a)t-a^2<0$$ is true for all $t\in[-1,1],$ for which we need
$$(-1)^2+(1-a)(-1)-a^2<0$$ and
$$1^2+(1-a)\cdot1-a^2<0,$$ which is
$$a^2+a-2>0$$ and $$a^2-a>0,$$ which gives
$$a\in(-\infty,-2)\cup(1,+\infty)$$ and we got the answer: $2$.