# Help showing $\sin(x)^{\sin(x)} < \cos(x)^{\cos(x)}$ between $x = 0$ and $x = \frac{\pi}{4}$?

This is a homework problem that I just can't make progress on. I know that it can be shown through a special case of the generalized arithmetic-geometric mean inequality ($$a^t b^{1-t} \leq ta + (1-t) b$$), but I just can't choose variables correctly. Hoping for a hint, not a solution. I've tried setting $$t = \sin^2 x$$, $$a = \sin x$$, $$b = \cos x$$ (so $$1-t = \cos^2 x$$), and several permutations of these, but I'm getting nowhere.

• Just prove that $\sin x < \cos x$ for $0 < x< \pi/4$. May 15 '19 at 1:04
• that's not enough since the powers go the other way, namely $(\sin x)^{\sin x} > (\sin x)^{\cos x}$ and same with $(\cos x)^{\sin x} >(\cos x)^{\cos x}$; note that at $0$ both terms converge to $1$, while at the other end again both terms are equal, so there can't be any simple monotonicity May 15 '19 at 1:08
• $$\sin x-\cos x=\sqrt2\sin(x-\pi/4)$$ and $$\sin x,\cos x>0$$ May 15 '19 at 1:12
• Take $f(x)=x^x$ and $f'(x)=x^x(1+\ln(x))<0$,. So $f(x)$ is increasing in $(0,e^{-1}]$.This means that $f(x)$ is both increasing and decreasing in .$[0,1/\sqrt{2}].$ So use of $x^x$ doesn't really help here and this makes the proof of Conard using a mix of AM-GM and calculus is valuable. May 15 '19 at 2:11

$$t=\frac{\sin x}{\cos x}, 1-t=\frac{\cos x-\sin x}{\cos x}, a=\sin x, b=\cos x+\sin x$$

Since $$0, we get $$0 1$$

$$a^t b^{1-t} \leq ta + (1-t) b$$ translates to

$$(\sin x)^{\frac{\sin x}{\cos x}}b^{1-t} \le \cos x$$.

Raising the above to power $$\cos x$$ and noting that $$b>1, 1-t>0$$, so $$b^{\frac{1-t}{\cos x}}>1$$, we get

$$(\sin x)^{\sin x}<(\sin x)^{\sin x}b^{\frac{1-t}{\cos x}} \le (\cos x)^{\cos x}$$, so done

Let $$f(x)=x^{1/\sqrt{1-x^2}}$$ for $$x \in (0,1).$$ $$\frac{f'(x)}{f(x)}=\frac{1-x^2+x^2 \ln x}{x (1-x^2)^{3/2}}$$ Now let $$g(x)=1-x^2+x^2 \ln x$$, $$g'(x)= x(\ln x^2-1)<0,$$ for $$x \in (0,1].$$ $$g(x)$$ is a decreasing function $$g(x) >g(1)=0$$, therefore $$f'(x)>0$$, $$f(x$$) is an increasing function. Finally, $$\sin x < \cos x,~ \mbox{for}~ x \in [0,\pi/4] \Rightarrow f(\sin x) < f(\cos x) \Rightarrow (\sin x) ^{\sec x} <(\cos x)^{\csc x} \Rightarrow (\sin x)^{\sin x} < (\cos x)^{\cos x}.$$

When $$x\in\left(0,\frac{\pi}{4}\right)$$, $$\sin x\in\left(0,\frac{1}{\sqrt{2}}\right),\ \tan x\in(0,1),\ \sin x+\cos x\in(1,\sqrt{2}).$$ It is easy to see that： $$(\sin x)^{\sin x}<(\cos x)^{\cos x} \iff(\sin x)^{^{\frac{\sin x}{\cos x}}}<\cos x.$$ Using Bernoulli's inequality$$(1+x)^{\alpha}<1+\alpha x$$ for $$0<\alpha<1$$ and $$-1, \begin{align*} (\sin x)^{^{\frac{\sin x}{\cos x}}} &=(1+(\sin x-1))^{^{\frac{\sin x}{\cos x}}}\\ &<1+\frac{\sin x}{\cos x}(\sin x-1)\\ &=\frac{\cos x+\sin^2 x-\sin x}{\cos x}. \end{align*} So we need only to prove that $$\frac{\cos x+\sin^2 x-\sin x}{\cos x}<\cos x \iff1<\cos x+\sin x.$$