Prove that $x+\frac1{4x} ≥ 1$ for $x>0$

Question

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:

1. $x = 1$
2. $x > 1$
3. $x < 1$

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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 $x > 0$.

Answer 1

$$x+\frac1{4x}-1=\frac{4x^2-4x+1}{4x}=\frac{(2x-1)^2}{4x}\ge0$$

Answer 2

by $AM-GM$ we have $$x+\frac{1}{4x}\geq 2\sqrt{x\cdot \frac{1}{4x}}=1$$

Answer 3

Another option, use Calculus to find the absolute minimum of $f(x) = x + \frac{1}{4x}$, and note that it is at least 1.

Answer 4

For $x\gt0$, $$ \begin{align} x-1+\frac1{4x} &=\left(\sqrt{x}-\frac1{2\sqrt{x}}\right)^2\\ &\ge0 \end{align} $$

Answer 5

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?).

Answer 6

An unusual way is letting $2x=\tan t>0$ then $$2x+\dfrac{1}{2x}=\dfrac{2}{\sin2t}\geq2$$

Answer 7

$$x+\frac1{4x}\ge 1 \iff 4x^2+1≥ 4x \iff4x^2-4x+1\ge 0\iff(2x-1)^2\ge 0$$