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I hope someone can help me giving a hint or sth for my inequality, which I'm trying to solve now for some days. I want to show that $$\frac{2}{\sqrt{\vphantom{\large A}1+c}}\ \leq\ \frac{1}{\sqrt{1+c\,\left(\frac{c\ +\ \sqrt{\vphantom{\Large A}c^{2}\ -\ 4\,}\,}{\vphantom{\Large A}2}\right)^{3}}} + \frac{1}{\sqrt{1+c\,\left(\frac{c\ -\ \sqrt{\vphantom{\Large A}c^{2}\ -\ 4\,}\,}{\vphantom{\Large A}2}\right)^{3}}}\,,\quad \forall\ c \geq 2$$

From plotting its easy to see and Mathematica/Maple also gave me the solution c>2, but thats not a real proof. Still from this, I think there must be a way to show it.

Besides just trying to show the inequality i tried showing that the difference of these terms is an injective function (for c=2 there is equality, which is a problem for many approximations). However the resulting terms dont really get easier to handle... Other than that I tried to use some meanvalue inequalities of harmonic mean, geometric mean, quadratic mean. The problem is that the inequalitiy between HM and GM is already too rough.

So at the moment im quite stuck and feel out of ideas how to approach the problem. It would be really nice if someone has some ideas (i hope i dont need a full solution :) )

P.S.: please excuse my bad english, I'm no native speaker.

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Have you tried to simplify right hand side? For example, let $x = \frac{c \pm \sqrt{c^2-4}}{2}$. Clearly, $x^2 - cx + 1 = 0$, so one can simplify $1+cx^3$ as $$1+cx(cx-1) = 1 - cx + c^2 (cx-1) = (c^2-1)(cx-1)$$ This shows that RHS has much room for simplification. –  user27126 Nov 2 '13 at 10:19

1 Answer 1

Let $c=e^x+e^{-x}$. Then $\sqrt{c^2-4} = e^x-e^{-x}$ so

$$1+c(\frac{c+\sqrt{c^2-4}}{2})^3=1+c(\frac{e^x+e^{-x}+e^x-e^{-x}}{2})^3=1+(e^x+e^{-x})e^{3x}=1+e^{2x}+e^{4x}$$

and

$$1+c(\frac{c-\sqrt{c^2-4}}{2})^3=1+c(\frac{e^x+e^{-x}-e^x+e^{-x}}{2})^3=1+(e^x+e^{-x})e^{-3x}=1+e^{-2x}+e^{-4x}=e^{-4x}(1+e^{2x}+e^{4x})$$

Then the right hand side is

$$\frac{1+e^{2x}}{\sqrt{1+e^{2x}+e^{4x}}}$$

Now to prove the inequality square both sides, cross multiply, and take the difference. The result is just a polynomial in $e^x$ and should factor nicely. Mathematica tells me it's

$$ e^{-x}(e^x-1)^4(1+e^x+e^{2x})$$

which is positive.

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You sir, are a sir of inequalities! As always its a question of the right substitution, so thanks for the good advice. I guess from the $\sqrt{c^{2}-4}$ one could have thought of something like $\cosh(x)$, but i guess my mind wasn't the clearest anymore. THANK YOU! –  user104857 Nov 2 '13 at 13:19
    
No problem, and yes, $\sqrt{c^2-4}$ hints at $\cosh(x)$. –  Prometheus Nov 3 '13 at 2:46

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