Prove this inequality holds $e^x+(\ln{x}-1)\sin{x}>0$

Question

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.

André Nicolas · Answer 1 · 2016-01-19T04:21:30.473

4

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 $[e,\infty)$.

edited Jan 19 '16 at 04:21

answered Jan 19 '16 at 03:29

André Nicolas

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Thanks,But How to prove $x>1$ case? – math110 Jan 19 '16 at 03:35
@qwr, But $(\ln{x}-1) \sin{x} $ can't known sign – math110 Jan 19 '16 at 03:38
In our interval, $\ln x$ is negative, so $(\ln x-1)\sin x$ is negative. That is why we need to do some work, in order to show that adding the $e^x$ makes it positive. – André Nicolas Jan 19 '16 at 03:44
I added a short paragraph on the easier $x\ge 1$ case. – André Nicolas Jan 19 '16 at 03:49
No, $1<x<e$case,then $\ln{x}-1<0$,in Fact,$e^x-(\ln{x}-1)\sin{x}<e^x-(\ln{x}-1)$ – math110 Jan 19 '16 at 03:57
My reply to your sign question was done before I added the little paragraph about $x\ge 1$. I will increase the length of the paragraph. – André Nicolas Jan 19 '16 at 04:02
I realized that it was simpler to just deal with two intervals, $(0,e)$ and the rest, separating out $[1,e)$ was a waste of time. – André Nicolas Jan 19 '16 at 04:23

score 2 · Accepted Answer · answered Mar 15 '17 at 10:55

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 $$(1-\ln x)x\le 1<e^x,\quad\forall x>0$$

score 0 · Answer 3 · answered Mar 14 '17 at 22:24

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.

Best regards.

Prove this inequality holds $e^x+(\ln{x}-1)\sin{x}>0$

3 Answers3