How to minimize $x+4+\frac{1}{2x-1}$?

Question

How can I minimize $x+4+\frac{1}{2x-1}$ for $x > \frac{1}{2}$? I have tried using AM-GM to get the inequality

$2\sqrt{\frac{x+4}{2x-1}} \leq x+4+\frac{1}{2x-1}$.

Then, using that AM = GM when the terms are equal, I got that $x = \frac{-7+\sqrt{89}}{4}$. However, when I used calculus, I got that the minimum had to be at $\frac{1+\sqrt{2}}{2}$.

I'm not sure where I went wrong. Can someone explain how to solve this without calculus?

Answer 1

Hint:   write it as $\;\dfrac{9}{2}+ \left(x-\dfrac{1}{2}\right)+\dfrac{1}{2x-1}\,$, first, then think AM-GM.

Answer 2

This is not an answer but an illustration.

I agree with dxiv that OP found a lower bound instead of greatest lower bound. Actually, when we split an expression into 2 and then use A.M.$\geq $G.M., we can create many lower bound, which may be dependent on a variable. In order to get the greatest lower bound a constant, we need to split the expression wisely as dxiv.

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