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I need to show that for $x \in (0, \pi/2)$

$$ e^{\cos x} \leq (e -1) \cos x + 1 $$

I was thinking on using extrema of some function, but i have no idea. please, help me. thanks

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3  
For all $x$? Or for restricted ranges? –  Babak S. Dec 11 '13 at 14:47
    
For what range? LHS is positive $\forall x \ \in \mathbb{R}$, but RHS is negative for a range of $x$ –  Alex Dec 11 '13 at 14:56
2  
sorry misread my problem, forgot to add x is in (0,pi) –  user113609 Dec 11 '13 at 14:57
    
Your modified range doesn't work either. The inequality does not hold for $x = 1.6 < \pi$. –  Macavity Dec 11 '13 at 14:58
    
Are you sure? You probably mean $\frac{\pi}{2}$ –  Alex Dec 11 '13 at 15:03

4 Answers 4

We may substitute $\cos x=:u$ and then have to prove that $$e^u\leq(1-u)1+u\>e\qquad(0\leq u\leq1)\ .$$ Since we have equality at $u=0$ and $u=1$ the claim immediately follows from the convexity of the exponential function.

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So simple an answer compared to the usual maxima/minima technique as I used in my answer. +1 –  Paramanand Singh Dec 11 '13 at 15:23

Let $f(x) = (e - 1)\cos x + 1 - e^{\cos x}$ so that $f(0) = 0 = f(\pi/2)$. Now $f'(x) = e^{\cos x}\sin x - (e - 1)\sin x = \sin x\{e^{\cos x} - e + 1\}$. Clearly we can see that $f'(x) = 0$ implies either $x = 0$ or $\cos x = \log (e - 1)$. Let $\alpha \in (0, \pi/2)$ such that $\cos \alpha = \log(e - 1)$. If $x < \alpha$ then $\cos x > \cos \alpha$ and hence $e^{\cos x} > e^{\cos \alpha} = e - 1$ so that $f'(x) > 0$ and similarly if $x > \alpha$ then $f'(x) < 0$. So the derivative changes sign from $+$ to $-$ as $x$ crosses $\alpha$. Therefore $x = \alpha$ gives a maximum of $f(x)$. It now follows that absolute minimum of $f(x)$ is attained at $x = 0$ or $x = \pi/2$ and absolute maximum is attained at $x = \alpha$. It follows that $f(x) \geq 0$ for all $x \in [0, \pi/2]$. Thus $e^{\cos x} \leq (e - 1)\cos x + 1$ for all $x \in [0, \pi/2]$.

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Notice: for any $x,y$ non-negative reals, $\alpha + \beta = 1 $, then

$$ x^{\alpha}y^{\beta} \leq \alpha x + \beta y $$

To show this, put $f(t) = (1 - \beta) + \beta t - t^{\beta} $. Show this function decreases on $[0,1]$ and then replace $t$ with $\frac{y}{x}$. Now, as an application of this with $x = e$, $\alpha = \cos x $, $y = 1$, $\beta = 1 - \cos x $, we get

$$ e^{\cos x} = e^{\cos x} 1^{1 - \cos x} \leq e \cos x + 1 ( 1 - \cos x ) $$

$$ \implies e^{\cos x } \leq e \cos x + 1 - \cos x = (e - 1) \cos x + 1 $$

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Assuming you meant $x \in [0 \frac{\pi}{2}]$. Consider $h(x)=(e-1) \cos x +1- e^{\cos x}$. Clearly $h(0)=0$ and $h(\frac{\pi}{2})=0$. Take the derivate. How does the derivative behave on the $[0, \frac{\pi}{2}$] segment? On this segment $h(x)$ has only one extreme point, which is local maximum. Before reaching this value, $h(x)$ monotonically grows, then monotonically decreases. What can you deduce from this?

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Our answers are the same but differing by a minute!! –  Paramanand Singh Dec 11 '13 at 15:20

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