Can This Trigonometric Inequality Be Proven for All Real Numbers?

[anemone]

Prove that \(\displaystyle \frac{\sin^3 x}{(1+\sin^2 x)^2}+\frac{\cos^3 x}{(1+\cos^2 x)^2}\lt \frac{3\sqrt{3}}{16}\) holds for all real $x$.

 

[Albert1]

Prove that \(\displaystyle \frac{\sin^3 x}{(1+\sin^2 x)^2}+\frac{\cos^3 x}{(1+\cos^2 x)^2}\lt \frac{3\sqrt{3}}{16}\) holds for all real $x$.

my solution:

Spoiler

let $f(x)=\dfrac{sin^3 x}{(1+sin^2x)^2}$
$g(x)=\dfrac{cos^3 x}{(1+cos^2x)^2}$
and $h(x)=f(x)+g(x)$
the maxmium of $h(x)$ will occur at some intersecion points of $f(x)$ and $g(x)$, and one of them is $x=\dfrac {\pi}{4}$
and $max(h(x))=2\times f(\dfrac {\pi}{4})=2\times g(\dfrac {\pi}{4})=\dfrac {2\sqrt 2}{9}<\dfrac {3\sqrt 3}{16}$ holds for all real $x$

 

[anemone]

Well done, Albert!

Hint for solving it with another method:

Spoiler

AM-GM inequality