Hi yesterday I see a question (closed) about this inequality :

Let $0<x\leq\frac{\pi}{4}$ then we have : $$\sin(x)^{\sin(x)}\leq \cos(x)^{\cos(x)}$$

I found that beautiful so I propose a proof without derivatives :

I recall that (for :$0<x\leq\frac{\pi}{4}$):

$$\cos(x)=\frac{1}{\sqrt{1+y^2}} \quad \sin(x)=\frac{y}{\sqrt{1+y^2}}$$

Where $y=\tan(x)$

Now we make the subsitution $y=\sinh(a)$ we have :

$$\cos(x)=\frac{1}{\cosh(a)} \quad \sin(x)=\tanh(a)$$

So we have :

Let $0<a\leq\sinh^{-1}(1)$ then we have : $$\tanh(a)^{\tanh(a)}\leq \Big(\frac{1}{\cosh(a)}\Big)^{\frac{1}{\cosh(a)}}$$

Now we use the Bernoulli's inequality to get : $$\tanh(a)^{\tanh(a)}\leq (1+\sinh(a)(\tanh(a)-1))^{\frac{1}{\cosh(a)}}$$

Recalling that (See comment of MartinR): $$\tanh(a)^{\tanh(a)}=\Big(1+\tanh(a)-1\Big)^{\tanh(a)}=\Big(1+\tanh(a)-1\Big)^{\frac{\sinh{x}}{\cosh(x)}}$$

So it sufficient to prove :

$$(1+\sinh(a)(\tanh(a)-1))^{\frac{1}{\cosh(a)}}\leq \Big(\frac{1}{\cosh(a)}\Big)^{\frac{1}{\cosh(a)}}$$

Or :

$$(1+\sinh(a)(\tanh(a)-1))\leq \Big(\frac{1}{\cosh(a)}\Big)$$

Multiplying by $\cosh(a)$ we get :

$$(\cosh(a)+\sinh(a)(\sinh(a)-\cosh(a)))\leq 1$$


$$(\cosh(a)+\sinh(a)(\sinh(a)-\cosh(a)))-1\leq 0$$

Putting $u=e^a$ and multiplying by $u^2$ we get a cubic polynomials wich can be easily factored since we have one as root .


My question :

Is it right ?

Have you an alternative proof ?

Thanks in advance for all your contributions .


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