Prove that tan &#x2061;<!-- ⁡ --> x &lt; 4 &#x03C0;<!-- π --> </mfra

Prove that
$\mathrm{tan}x<\frac{4}{\pi }x,\mathrm{\forall }x\in \left(0;\frac{\pi }{4}\right)$
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Elias Flores
Let g be the function defined on the interval $\left[0,\pi /4\right]$ as
$g\left(x\right)=\left\{\begin{array}{ll}\frac{\mathrm{tan}x}{x}& ,x\ne 0\\ \\ 1& ,x=0\end{array}$
Then, the derivative g′ of g is given by
${g}^{\prime }\left(x\right)=\frac{x{\mathrm{sec}}^{2}x-\mathrm{tan}x}{{x}^{2}}=\frac{x-\frac{1}{2}\mathrm{sin}\left(2x\right)}{{x}^{2}{\mathrm{cos}}^{2}x}>0$
for x>0 and g′(0)=0.
Inasmuch as g is increasing for $x\in \left[0,\pi /4\right]$ it attains, therefore, its maximum there at $x=\pi /4$
Thus,
$g\left(x\right)
and we are done!