<s>[INST] Demonstrate inequality $$15\le(3+\sin^2x)(4+\cos^2x)\le16 \mbox{ for any }x \in \mathbb{R}$$
I've wrote everything using $\sin$:
\begin{align*}15\le(3+\sin^2x)(4+1-\sin^2x)\le16&\Rightarrow
15\le(3+\sin^2x)(5-\sin^2x)\le16\\ 
&\Rightarrow 15\le15+2\sin^2x-\sin^4x\le16 \mid-15\\
&\Rightarrow 0\le2\sin^2x-\sin^4x\le1.
\end{align*}
I've stuck here and I need some help. Thanks.
 [/INST] Welcome to math.stackexchange Daniel!
Just expand and simplify the middle term as such: $$ (3 + \sin^2 x)(4+\cos^2 x) = 12 + 3\cos^2 x + 4\sin^2 x + \sin^2 x \cos^2 x $$
Now, since $\sin^2 x + \cos^2 x = 1$ we can write that term as $$ 15 + \sin^2x + \sin^2 x\cos^2 x$$
Thus, this reduces our problem to showing 
$$ 0 \leq \sin^2 x + \sin^2 x \cos^2 x \leq 1$$
which is easy to see if we write that term as $$ (1-\cos^2 x) + \sin^2 x \cos^2 x = 1+ \cos^2 x(\sin^2x -1) = 1-\cos^4 x.$$

As Srivatsan pointed out, a quicker route is to change into cosines immediately:
$$ (3+\sin^2 x)(4+\cos^2 x) = (3 + (1-\cos^2 x))(4+\cos^2 x) $$
$$ = (4-\cos^2 x)(4+\cos^2 x) = 16 - \cos^4 x$$
which produces the required inequality since $0\leq \cos^4 x \leq 1$.
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