0
$\begingroup$

Hello I have a problem with an inequality :

let $x$ a positive real number we have :

$0\geq -[\cos(\arctan((7x)^{\frac{1}{2}}))^{\frac{2}{3}}+ \cos(\arctan((\frac{7}{x})^{\frac{1}{2}})))^{\frac{2}{3}}] + \cos(\arctan(\sqrt{3x})) + \cos(\arctan(\sqrt{\frac{3}{x}}))$

I have a no idea to prove that properly .

Edit : At the beginning I want to prove this inequality from this link

So my idea was to use trigonometric identites like this :

$\cos(\arctan(\sqrt{\frac{3a}{b}}))=\sqrt{\frac{b}{3a+b}}$

And

$\cos(\arctan((\frac{7a}{b})^{\frac{1}{2}}))^{\frac{2}{3}}=(\frac{b}{7a+b})^{\frac{1}{3}}$

My second idea is to make the classical substitution :

$a=bx$

And now it gives the inequality you have above .

To prove that I think we can use this and this

Thanks

$\endgroup$
  • $\begingroup$ I think you're missing a closing bracket somewhere. $\endgroup$ – aardvark2012 Aug 12 '17 at 11:39
0
$\begingroup$

Hint: $$\cos \left(\arctan \sqrt{3x}\right)+\cos \left(\arctan \sqrt{3/x}\right)-\cos \left(\arctan \sqrt{7x}\right)^{2/3}-\cos \left(\arctan \sqrt{7/x}\right)^{2/3} \\ = \frac{1}{(3x+1)^{1/2}}+\frac{1}{(3/x+1)^{1/2}}-\frac{1}{\left(7x+1\right)^{\frac{1}{3}}}-\frac{1}{\left(7/x+1\right)^{\frac{1}{3}}}$$ Just note that $\cos(\arctan(f)) = \frac{1}{\sqrt{f^2+1}}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy