MHB Can This Trigonometric Inequality Be Proven for All Real Numbers?

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The discussion centers on proving the inequality $$\frac{\sin^3 x}{(1+\sin^2 x)^2}+\frac{\cos^3 x}{(1+\cos^2 x)^2}\lt \frac{3\sqrt{3}}{16}$$ for all real numbers x. Participants explore various approaches to establish the validity of the inequality, with one member providing a solution. Additional hints are offered for alternative methods of proof. The conversation emphasizes the mathematical rigor required to tackle trigonometric inequalities. The inequality is presented as a challenge for further exploration and validation.
anemone
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Prove that $$\frac{\sin^3 x}{(1+\sin^2 x)^2}+\frac{\cos^3 x}{(1+\cos^2 x)^2}\lt \frac{3\sqrt{3}}{16}$$ holds for all real $x$.
 
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anemone said:
Prove that $$\frac{\sin^3 x}{(1+\sin^2 x)^2}+\frac{\cos^3 x}{(1+\cos^2 x)^2}\lt \frac{3\sqrt{3}}{16}$$ holds for all real $x$.
my solution:
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 $x$
 
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Well done, Albert!

Hint for solving it with another method:

AM-GM inequality
 
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