Solution to : $g(x)=(\sin^2(x))^{\frac{1}{f(x)}}+(\cos^2(x))^{f(x)}\leq 1$

Question

Well, this is related to my own question Pretty conjecture $x^{\left(\frac{y}{x}\right)^n}+y^{\left(\frac{x}{y}\right)^n}\leq 1$ .

Let $x\in(-\infty,\infty)$ and define $g(x),f(x)$ continuous and differentiable functions such that : $$g(x)=(\sin^2(x))^{\frac{1}{f(x)}}+(\cos^2(x))^{f(x)}\leq 1\quad (1)$$

Is the function $f(x)=(\tan^2(x))^y$ (where $y$ is a positive or negative constant or equal to zero) the only solution to the inequality $(1)$ ?

Almost all of the elementary functions I came across alternated around $1$ so :

Do you have an example ?

Thanks a lot for all your contribution.

Answer 1

Let$x\in (-{\infty},{\infty})$ we put $g(x)=(\sin(x)^2)^{\frac{1}{f(x)}}+(\cos(x)^{2})^{f(x)}\leq(1)$ so $g(x)\leq\cos(x)^2+\sin(x)^2$ From it we find the following asymmetry$\cos(x)^2(1-(\cos(x)^2)^{f(x)-1})+(\sin(x)^2)(1-(\sin(x)^2)^{\frac{1-f(x)}{f(x)}}\geq1\implies\{_{(\cos(x)^2)^{f(x)-1}\leq0}^{(\sin(x)^2)^{\frac{1-f(x)}{f(x)}}\leq0}\implies\{_{f(x)-1=0}^{\frac{1-f(x)}{f(x)}=0}\implies (f(x)=1)$In the end we find that $g(x)\geq(1)$ Realized for functions equal to one