Prove :
$e^{\sin^{2} x}-e^{\cos^{2} x}≤|\cos (2x)|(e-1)$
I don't know how I prove it but my try as following : We know that :
$|a|-|b|≤|a-b|≤|a|+|b|$ so we obtain :
$e^{\cos^{2} x}- e^{\sin^{2} x}≤e^{\sin x}+e^{\cos x}$ But I don't have any ideas to approach it ?
If any one can give me some hints? Thanks
On
Define the function
$$f(x) = (e-1)|\cos 2x| -(e^{\sin^2x} - e^{\cos^2 x}) $$
and reexpress it as follows,
$$f(x) = 2e^{\frac 12} \left[|\cos 2x|\sinh\frac 12 +\sinh\left( \frac12 \cos 2x\right)\right]$$
or, with $-1\le t \le 1$,
$$f(t) = 2e^{\frac 12} \left(|t|\sinh\frac 12 +\sinh \frac t2\right)$$
Note that $f(t) \ge 0$ over $[0,1]$. For $[-1,0]$, we verify that its second derivative is negative,
$$f''(t) = \frac {e^{\frac 12}}{2}\sinh \frac t2 \le 0$$
Also, verify that $f(0) =f(-1) = 0$. That is, $f(t)$ is a concave function over $[-1,0]$ with boundary values at zero. Thus, $f(t)\ge 0$ and, hence, the original inequality,
$$ e^{\sin^2x} - e^{\cos^2 x} \le (e-1)|\cos 2x| $$
On
Let $$f(x)= \frac{e^{\cos^2 x}- e^{\sin^2 x}}{\cos 2x}$$ Thios function is even periodic woth period $\pi/2$. $$f'(x)=sec 2x \tan 2x [-2 e^{\cos^2 x}+2 e^{\sin^2 x}+ (e^{\cos^2 x} + e^{\sin^2 x}) \cos 2x]$$ in $[0,\pi/2]$ its roots are $x=\pi/4, \pi/2$. Next $f(\pi/2)=e-1$ and $f(\pi/4)=\frac{0}{0}$ so by taking the limit of $f(x)$ as $x \rightarrow \pi/4$ by L'Hospital rule, we get $$\lim_{x \rightarrow \pi/4} \frac{-\sin 2x e^{\cos^2 x}-\sin 2x e^{\cos^2 x}}{-2\sin 2x}=\sqrt{e}.$$ So the max of the function is $(e-1)$ and its minimum is $\sqrt{e}.$
If we write $u=\cos{2x}$, then the double-angle formulae give $$ \sin^2{x} = \frac{1-u}{2}, \quad \cos^2{x} = \frac{1+u}{2} $$ So we need to understand what happens to the function $$ e^{\cos^2{x}} - e^{\sin^2{x}} = e^{1/2+u/2} - e^{1/2 - u/2} = 2\sqrt{e} \sinh{(u/2)} $$ as $u$ varies between $-1$ and $1$. This function is odd, so it will suffice to consider $u>0$, then what happens to negative $u$ will follow easily. In particular, $\sinh{(u/2)}$ is convex for positive $u$ (its second derivative is positive), so it lies below the chord joining $(0,0)$ to $(U,\sinh{(U/2)})$ on the interval $[0,U]$. Choosing $U=1$, this chord has the equation $ Y = (e-1)X $, so for $0 \leq u \leq 1$, $$ 2\sqrt{e} \sinh{(u/2)} \leq (e-1)u . $$
Both sides of this are odd functions, so if $-1 \leq u \leq 1$, then $$ \lvert 2\sqrt{e} \sinh{(u/2)} \rvert \leq (e-1) \lvert u \rvert . $$ Replacing $u$ by $\cos{2x}$ again then gives $$ \lvert e^{\cos^2{x}} - e^{\sin^2{x}} \rvert \leq (e-1) \lvert \cos{2x} \rvert , $$ as required.