Prove that $x+\frac1{4x} ≥ 1$ for $x>0$ 
Let $x$ be a real number such that $x > 0$. Prove that x +$\frac {1} {4x}  ≥ 1$.

Not really sure on the correct way to approach it/is valid and could use some help.
Answer:
Proof Strategy: Proof by cases:


*

*$x = 1$

*$x > 1$

*$x < 1$


--
Case 1: $x = 1$
$x+\frac1{4x}$
$= (1) + 1/(4(1))$
$= 1.25 \ge 1$
Case is true
Case 2: $x > 1$, in this case $x = 2$.
$x+\frac1{4x} $
$= (2) + 1/(4(2))$
$= 2.125 \ge 1$
Case is true
Case 3: $x < 1$, in this case $x = 0.5$.
$= (0.5) + 1/(4(0.5))$
$= 1  \ge 1$
Case is true
Since all possible cases were satisfied therefore $x+\frac1{4x} \ge 1 $ when $x > 0$.

For fun:
Let $x>0,$ real. Multiply by $4x$ :
$4x^2-4x +1 \ge 0.$
Need to show that above inequality is true for $x \gt 0$.
$4x^2 -4x +1 = 4(x^2 -x)  +1 = 4(x-1/2)^2  \ge 0$
(why?).
