Find the greatest $a \in \mathbb{Z}$ such that $x^2-ax-\ln x+e^{\sin x}-1>0$ holding for every $x>0$. 
Find the greatest $a \in \mathbb{Z}$ such that  $$x^2-ax-\ln x+e^{\sin
 x}-1>0$$ holding for every $x>0$. (Reference data: $\sin 1 \approx 0.84, \ln2 \approx 0.693)$

Notice that the inequality is equivalent to
$$f(x):=\frac{x^2-\ln x+e^{\sin x}-1}{x}>a.$$
If we can find the globle minimum of $f(x)$, then the problem is solved. But $f(x)$ is a transcendent function, it's not so easy to obtain the exact minimum. How to enlarge or contract?
 A: Answer: greatest $a=2$. Proof.
If $a=3$ and $x=e$ then $x^2-ax-\ln x+e^{\sin x}-1<0$. 
Let $f(x)=x^2-2x-\ln x+e^{\sin x}-1$. We need to prove that $f(x)>0$ for all $x$.
Cases:
I. Let $x \ge e$. Let $g(x)=x^2-2x-\ln x-1$. Then $g'(x)=2x-2-\frac{1}{x}=\frac{2}{x} \left( x-\frac{\sqrt3+1}{2}\right)\left( x+\frac{\sqrt3-1}{2}\right)$. Since $e>\frac{\sqrt3+1}{2}$ then $\min \limits_{x \ge e} g(x)=g(e)$. Then $f(x)=g(x)+e^{\sin x}\ge e^2-2e-2+e^{\sin x}\ge e^2-2e-2+e^{-1}>0$.
II. Let $\sqrt2+1 \le x < e$. Then $x^2-2x-1=(x-1)^2-2 \ge 0$ and $\ln x<1$ and $e^{\sin x}>1$ (because $e<\pi$, $\sin x>0$). Then $f(x)>0$.
III. Let $\pi-1 \le x < \sqrt2+1$. Then $e^{\sin x}\ge e^{\sin(\sqrt2+1)}>1.94$. Then $f(x)=x(x-2)-\ln x+e^{\sin x}-1>(\pi-1)(\pi-3)-\ln(\sqrt2+1)+1.94-1>0$.
IV. Let $1 \le x < \pi-1$. Then $e^{\sin x} \ge e^{\sin 1}>2.3$. Then $f(x)>x^2-2x-\ln x+1.3$. Let $g(x)=x^2-2x-\ln x+1.3$.  Then $g'(x)=2x-2-\frac{1}{x}=\frac{2}{x} \left( x-\frac{\sqrt3+1}{2}\right)\left( x+\frac{\sqrt3-1}{2}\right)$.Then $\min \limits_{1 \le x<\pi-1} g(x)=g(\frac{\sqrt3+1}{2})=\left( \frac{\sqrt3+1}{2}\right)^2-2\left( \frac{\sqrt3+1}{2}\right)-\ln \left( \frac{\sqrt3+1}{2}\right)+1.3>0$. Then $f(x)>g(x)>0$.
V. Let $\frac{\pi}{4} \le x <1$. Then $e^{\sin x} \ge e^{\sin \frac{\pi}{4}}>2$. Then $e^{\sin x}-2-\ln x >0$. Then $f(x)=(x-1)^2+e^{\sin x}-2-\ln x >0$.
VI. Let $0<x<\frac{\pi}{4}$. It hard case. $$f'(x)=2x-2-\frac{1}{x}+\cos x \cdot e^{\sin x}$$ $$f''(x)=2+\frac{1}{x^2}-\sin x \cdot e^{\sin x}+\cos^2 x \cdot e^{\sin x}$$ $$f'''(x)=-\frac{2}{x^3}-\cos x \sin^2 x \cdot e^{\sin x}-3\sin x \cos x \cdot e^{\sin x}$$ Ease to see that $f'''(x)<0$. Then $\min \limits_{0<x<\frac{\pi}{4}} f''(x)=f''(\frac{\pi}{4})=2+\frac{16}{\pi^2}-\frac{\sqrt2}{2}e^{\frac{\sqrt2}{2}}+0.5e^{\frac{\sqrt2}{2}}>0$. Then $f''(x)>0$. Then $\max \limits_{0<x<\frac{\pi}{4}} f'(x)=f'(\frac{\pi}{4})=\frac{\pi}{2}-2-\frac{4}{\pi}+\frac{\sqrt2}{2}e^{\frac{\sqrt2}{2}}<0$. Then $f'(x)<0$. Then $\min \limits_{0<x<\frac{\pi}{4}} f(x)=f(\frac{\pi}{4})=\left(\frac{\pi}{4} \right)^2-\frac{\pi}{2}-\ln \frac{\pi}{4}+e^{\frac{\sqrt2}{2}}-1>0$.
Proof is complete.
