Find the image of function $f$ (without calculus)

Question

We have the function $f:R->R$, $f(x)=sin^{2n}(x)+cos^{2n}(x)$, where $n\geq1$ natural number. Find the image of $f$. I did the problem with calculus and the image is $[2^{1-n},1]$, but I want to solve it only using inequalities. I used AM-GM inequality I got $f(x)\geq 2^{1-n}sin^{n}(2x)$. We can see that the equality holds when $x=\pi/4$. The idea with AM-GM looks too good.

Answer 1

Since $n\geq1$, we obtain: $$f(x)\leq\sin^2x+\cos^2x=1.$$ The equality occurs for $x=0$, which says that we got a maximal value.

Also, for $n\geq2$ by Holder $$f(x)=\frac{(1+1)^{\frac{n}{2}-1}\left(\sin^{2n}x+\cos^{2n}x\right)}{2^{\frac{n}{2}-1}}\geq\frac{\left(\sin^4{x}+\cos^4x\right)^{\frac{n}{2}}}{2^{\frac{n}{2}-1}}\geq\frac{\left(\frac{1}{2}\right)^{\frac{n}{2}}}{2^{\frac{n}{2}-1}}=2^{1-n}.$$ The equality occurs for $x=45^{\circ},$ which says that we got a minimal value.

Answer 2

HINT AND COMMENT.- Rationalizing with the tangent, $t$, we have $$\sin^2(x)=\frac{t^2}{1+t^2}; \cos^2(x)=\frac{1}{1+t^2}$$ so we have to bound the function $$f(t)=\frac{t^{2n}+1}{(t^2+1)^n}$$Note that $f(0)=1$ and $f(1)=\dfrac{1}{2^{n-1}}$ and prove that $$\dfrac{1}{2^{n-1}}\le f(t)\le1$$ The part $f(t)\le1$ is obvious and only you have to prove $\dfrac{1}{2^{n-1}}\le f(t)$.