Prove this inequality holds $e^x+(\ln{x}-1)\sin{x}>0$ Let $x>0$.show that this following inequality
$$e^x+(\ln{x}-1)\sin{x}>0$$
I tried doing this with derivatives, but I don't quickly found that it was outside of my ability to obtain the necessary derivatives, so I figured there must be some simpler way to do this, but I don't really know how.
 A: Outline: We first look at  $x$ in the interval $(0,e)$. Note that $e^x\gt 1+x$ and $0\lt \sin x\lt x$. So it will be enough to show that $1+x+(\ln x-1)x\gt 0$ in the interval.  
Now we can use the derivative. Let $g(x)=1+x\ln x$.  Show that the minimum value of $g(x)$ in the interval $(0,e)$ is positive.
The interval $[e,\infty)$ is easier to deal with.  Because $\sin x\ge -1$, on the interval $[e,\infty)$ our function is $\ge e^x-(\ln x-1)$, which is positive at $e$ and increasing in $[e,\infty)$.
A: The existing answers are cool, but you may want to consider another approach as well.
In order to prove that
$$e^x+(\ln{x}-1)\sin{x}>0$$
it is sufficient to prove
$$e^x>(1-\ln x)x$$
due to the fact that $x>\sin x$ for all positive $x$. Since $\ln x$ is strictly increasing and $\ln 1=0$ then it is safe to say that
$$\int_x^0\ln t\,dt\le\int_1^0\ln t\,dt$$
Ironically, the left-side of the above inequality is $(1-\ln x)x$ and the right side is $1$. Therefore
$$(1-\ln x)x\le 1<e^x,\quad\forall x>0$$
A: Notice that $\ln{x}\leq x-1$ so $e^x+(\ln{x}-1)\sin{x}\geq e^x - (2-x)\sin x$. On the other hand, $-ax \leq a\sin(x) \leq ax$ with $a \in \mathbb{R}^+$, so that $e^x - 2\sin x + x \sin x \geq e^x + 2x - x^2$. i.e: is enough to show $e^x - (x^2 - 2x) > 0$ but $e^x - (x^2 - 2x) = e^x - (x-1)^2 + 1 > e^x - (x-1)^2 $ and since $e^x > 1$ then $e^x - (x-1)^2$ > 0.
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