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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. $x < 1$

--

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 $x > 0$.

iSNSD09
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    In case 2 you don't know $x = 2$. You must write an argument that works for any $x > 1$. That can be done, and so yours is an OK strategy, even if not the best. (See the answers.) – Ethan Bolker Feb 06 '18 at 16:36
  • Editors: Be careful not to change the intent of the question when editing – robjohn Feb 06 '18 at 17:07
  • @iSSNSD09: You have shown the result is true for $3$ particular values, not for all values. – robjohn Feb 06 '18 at 17:13
  • The question is tagged [proof-verification] and asks if the given proof is valid. I cannot see that any of the answers given so far addresses this question, they all provide alternate proofs. – If OP were just asking for some proof then we could simply close as a duplicate of Inequality $x + \frac1{4x} \ge 1$ holds for all $x > 0$ – Martin R Feb 06 '18 at 21:21
  • @EthanBolker how would I make an argument to show that for any x > 1? – iSNSD09 Feb 07 '18 at 18:07
  • @MartinR You're correct, I don't think anyone has provided the answer that I'm looking for yet. I think the solution is much simpler than how everyone is approaching it at. I looked at the question you linked however I haven't learned anything on AM–GM so I don't think this would apply to finding the solution. – iSNSD09 Feb 07 '18 at 18:09
  • @iSNSD09 Please, if you are ok, you can accept the answer and set it as solved. Thanks! cdn.sstatic.net/img/faq/faq-accept-answer.png – user Feb 08 '18 at 22:26

7 Answers7

8

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

Roman83
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  • Roman.Very nice! – Peter Szilas Feb 06 '18 at 17:00
  • @PeterSzilas: Thank you! – Roman83 Feb 06 '18 at 17:02
  • How did you subtract the - 1 given the right side of the intiall equation doesn't equal to one. Also where does multiplying everything by 4x come from? Now that we have ((2x-1)^2)/4x >= 0, how would you then prove that that is bigger than 0? – iSNSD09 Feb 07 '18 at 18:04
  • @iSNSD09:

    1. $$x+\frac1{4x}\ge1$$ if only if $$x+\frac1{4x}-1\ge0$$;

    2. $x>0$ and $(2x-1)^2\ge 0$ then $$\frac{(2x-1)^2}{4x}\ge0$$

     – Roman83 Feb 07 '18 at 20:12
5

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

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

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

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

robjohn
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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?).

Peter Szilas
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0

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

Nosrati
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