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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 prove each side of the inequality by using the fact that $\sin^2 x \leq 1$ in two different ways. Let me start with the left side of the inequality. $2 \sin^2 x - \sin^4 x = (\sin^2 x) \cdot (1 + (1 - \sin^2 x))$. Can you see why this is nonnegative? –  Srivatsan Sep 20 '11 at 12:55
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(Another approach to try is to write down everything in terms of $\cos$ instead of $\sin$.) –  Srivatsan Sep 20 '11 at 12:59
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Put $t=\sin^2 x$, then $0\le t\le1$ and $2\sin^2 x- \sin^4 x= 2t-t^2 = t(2-t)$ which reaches maximum at ... because ... –  AD. Sep 20 '11 at 13:07
    
@Srivatsan Narayanan: Love your last comment :D –  AD. Sep 20 '11 at 13:07
    
+1 for showing some effort. Others please note how this gets a more useful answer. –  Ross Millikan Sep 20 '11 at 13:23

2 Answers 2

up vote 3 down vote accepted

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 $[15, 16]$. –  Srivatsan Sep 20 '11 at 13:05

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! –  Daniel Sep 20 '11 at 13:27

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