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.

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    $\begingroup$ Just prove that $\sin x < \cos x$ for $0 < x< \pi/4$. $\endgroup$ May 15 '19 at 1:04
  • $\begingroup$ 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 $\endgroup$
    – Conrad
    May 15 '19 at 1:08
  • $\begingroup$ $$\sin x-\cos x=\sqrt2\sin(x-\pi/4)$$ and $$\sin x,\cos x>0$$ $\endgroup$ May 15 '19 at 1:12
  • $\begingroup$ 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. $\endgroup$
    – Z Ahmed
    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 $0<x<\frac{\pi}{4}$, we get $0<t<1, b=\frac{2}{\sqrt 2}\cos (\frac{\pi}{4}-x) > 1$

$a^t b^{1-t} \leq ta + (1-t) b$ 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 $-1<x\neq0$, \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 $$\frac{\cos x+\sin^2 x-\sin x}{\cos x}<\cos x \iff1<\cos x+\sin x.$$


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