After squaring of the both sides we need to prove that $$\sin{x}(4\sin^3x-3\sin{x}+4\sin^2x-3-1)<0,$$which is obvious.

Just differentiate and see the nature of function $f(x)=sin2x+cosx$

WolfusA wrote: Prove that for all $x \in \left( 0, \frac{\pi}{3} \right)$ inequality $sin2x+cosx>1$ holds. $sin2x+cosx>1\iff 2cosx>tan\frac{x}{2}$

WolfusA wrote: Prove that for all $x \in \left( 0, \frac{\pi}{3} \right)$ inequality $sin2x+cosx>1$ holds. Since $x\in \left( 0, \frac{\pi}{3} \right)$ $1>cosx >\frac12, sin 2x=2\sin x \cos x>sin x$ and $sin x + cos x >1 , x\in \left(0,\frac {\pi}{2}\right)$ by triangular inequality. done.

sqing wrote: WolfusA wrote: Prove that for all $x \in \left( 0, \frac{\pi}{3} \right)$ inequality $sin2x+cosx>1$ holds. $sin2x+cosx>1\iff 2cosx>tan\frac{x}{2}$ Can you show how did you get this? Because the forward part is very easy, and I can't reach inequality of yours.