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Prove the inequality:

$$\sin{x}<\frac{x}{\sqrt{1+\frac{x^2}{3}}}$$ for real $x>0$.

Any hints?

EDIT: Using calculus and derivatives is perfectly allowed.

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    from where does this inequality come? – 2017-02-23
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    Hint: The expressions are equal when $x=0$. If you are not allowed to use calculus (in particular, by comparing derivatives), then the only obvious alternative is geometry. – 2017-02-23
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    It is straightforward to verify that Maclaurin according to $x=0$ of both sides are equal up to $x^3$. A plot shows they are very close, and the hypothesis seems to be right. As pointed out, $x>\sqrt{3/2}$ is trivial. For the rest, I tried to introduce a remainder term, but was unable to bound it in a manner valid to all intervals. I split into many cases of intervals, but just got the restraint more cumbersome, and gave it up. – 2017-02-23
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    @Aminopterin: To keep the expressions manageable, consider tackling $\sin^2 x \lt x^2/(1+x^2/3)$ instead (for $x\gt 0$). – 2017-02-23
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    @user418969 Nice problem! – 2017-02-23

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This is an incomplete answer.

Perhaps you can do it easily for $x>\sqrt{\frac{3}{2}}$

$1+\frac{x^2}{3}\sqrt{\frac{3}{2}}$

Then

$(sin^2x+cos^2x)(1+\frac{x^2}{3})

Therefore

$(sin^2x)(1+\frac{x^2}{3})<(sin^2x+cos^2x)(1+\frac{x^2}{3})

$(sinx)\sqrt{(1+\frac{x^2}{3})}

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    This is the easy part, because $\frac{x}{\sqrt{1+\frac{x^2}{3}}>1\ge\sinx$ for $x>\sqrt{\frac32}$ :-) – 2017-02-23