1
$\begingroup$

Can someone help me show that $\cosh \geq1$.

I know that $\cosh x = \frac{1}{2}(e^x+s^{-x})$

I think I'm supposed to use the identity: $\cosh^2x - \sinh^2x=1$

  • 0
    Try to differentiate the expression for $coshx$ and you see that its minimum value is 1. – 2017-02-19

2 Answers 2

3

Note $$\cosh^2 x =1+\sinh^2 x \ge 1$$ So $\cosh^2 x \ge 1$. Since $\cosh x$ is positive, we have $$\cosh x \ge 1$$ Done! You could also say $$\cosh x =\frac{e^x+e^{-x}}{2} \ge \sqrt{e^{x} \times e^{-x}}=1$$ From $\text{AM-GM}$.

  • 0
    I tried this: http://mathb.in/127894 , but I was told that I assume what I try to prove. – 2017-02-19
  • 0
    @tonytouch So you aren't given $\cosh^2 x -\sinh^2 x=1$? – 2017-02-19
  • 0
    Yes, I'm allowed to use that. – 2017-02-19
  • 1
    @tonytouch Is my edit sufficent? – 2017-02-19
  • 0
    Yes, thank you so much! I can first accept your answer in 4 min! but the link I showed you, was that wrong? I posted it on an irc forum and I was told it was wrong. – 2017-02-19
  • 1
    @tonytouch No, I think that it should be right. Perhaps you should be elaborating on how $$\cosh^2x-\sinh^2x=1$$ And why $\cosh x>0$, as $$\cosh^2 x \ge 1$$ merely implies $\cosh x \le -1$ or $\cosh x \ge 1$. – 2017-02-19
1

by AM-GM we get : $$\frac{1}{2}(e^x+e^{-x})\geq \sqrt{e^x\cdot e^{-x}}=1$$

  • 0
    Thank you for your fast answer Dr. Sonnhard – 2017-02-19