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$15\le(3+\sin^2x)(4+\cos^2x)\le16 \mbox{ for any }x \in \mathbb{R}$

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.

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    You can also switch to cosines at the end: $2\sin^2 x - \sin^4 x = \sin^2 x (2 - \sin^2 x) = (1 - \cos^2 x)(1 + \cos^2 x) = 1 - \cos^4 x \in [0,1]$. But Sri's suggestion to switch to cosines right away seems easiest here.2011-09-20

2 Answers 2

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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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    Another way is to rewrite the equation in terms of $\cos$: $ (3+\sin^2 x) (4+\cos^2 x) = (4 - \cos^2 x) (4 + \cos^2 x) = 16 - \cos^4 x, $ which is clearly in $[15, 16]$.2011-09-20
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There are slightly shorter ways for this specific problem (see Ragib's answer and my comment below it), but I will instead push the OP's attempt to completion.

So, we are left with showing two inequalities: $ 2 \sin^2 x - \sin^4 x \geq 0, \tag{1} $ $ 2 \sin^2 x - \sin^4 x \leq 1. \tag{2} $

For (1): An important fact worth remembering is that $0 \leq \sin^2 x \leq 1$ for all real $x$. Multiplying by the nonnegative number $\sin^2 x$, we get $ \sin^4 x \leq \sin^2 x .$
You should be able to prove $(1)$ by plugging in this inequality. Can you complete this part?

For (2): For such inequalities, it is a good idea to collect terms together and see if the resulting expression can be simplified (by grouping terms or factoring). Collecting the terms on the right, the inequality $(2)$ is equivalent to: $ 1 - 2 \sin^2 x + \sin^4 x\geq 0 .\tag{3} $ You should be able to show this by factoring the left hand side. Can you take it from here?

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    That's a nice solution too. Thank you!2011-09-20