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This question already has an answer here:

Suppose $x > 0$. Prove that $$\frac{\sqrt{x}}{x+1} \leq \frac{1}{2}$$

Hey everyone, here is a simple math prove question, but I had a hard to start this proving, please give me some ideas to deal with this. Thanks

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marked as duplicate by JMoravitz, dxiv, Namaste, Arnaldo, Simply Beautiful Art May 11 '17 at 21:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Are you familiar with the concept of the derivative in calculus? $\endgroup$ – Franklin Pezzuti Dyer May 11 '17 at 20:40
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    $\begingroup$ Given $x>0$ one has $\frac{\sqrt{x}}{x+1}\leq \frac{1}{2}$ if and only if $2\sqrt{x}\leq x+1$ If and only if $4x \leq x^2+2x+1$ if and only if ... $\endgroup$ – JMoravitz May 11 '17 at 20:40
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If $x>0$ then

$$\frac{\sqrt{x}}{x+1} \leq \frac{1}{2}\Leftrightarrow 2\sqrt{x}\le x+1\Leftrightarrow x-2\sqrt{x}+1\ge 0\Leftrightarrow (\sqrt{x}-1)^2\ge 0$$

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Start by assuming that there is some integer $a$ for which $$\frac{\sqrt x}{x+1} \gt \frac{1}{2}$$ so that for some other positive number $b$, $$\frac{\sqrt x}{x+1} = \frac{1}{2}+b$$ Then we can proceed using algebra to elicit a contradiction: $${\sqrt x} = (\frac{1}{2}+b)(x+1)$$ $$x = (\frac{1}{2}+b)^2(x+1)^2$$ Let us set the quantity $(\frac{1}{2}+b)^2$ equal to $c$. Then $c$ is also a positive number that is greater than $\frac{1}{4}$, and $$x = c(x+1)^2$$ $$x = c(x^2+2x+1)$$ $$cx^2+(2c-1)x+c=0$$ Now we use the quadratic formula: $$x=\frac{1-2c\pm \sqrt{(2c-1)^2-4c^2}}{2c}$$ $$x=\frac{1-2c\pm \sqrt{4c^2-4c+1-4c^2}}{2c}$$ $$x=\frac{1-2c\pm \sqrt{1-4c}}{2c}$$ However, since $c$ is greater than $\frac{1}{4}$, then $1-4c$ is negative and $\sqrt{1-4c}$ is imaginary, showing us that there are no possible values of $x$ satisfying this.

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For $x=1$, $\frac{x}{(x+1)^2}=\frac{1}{4}$.

For $x\geq1$, the denominator of $\frac{x}{(x+1)^2}$ increases faster than the numerator so the fraction is decreasing. So for $x\geq1$, $\frac{x}{(x+1)^2}\leq\frac{1}{4}$.

For $0\leq x\leq1$, $x=\frac{1}{X}$ for some $X\geq1$. So then $\frac{x^2}{(x+1)^2}=\frac{1/X^2}{(1/X^2+1)^2}=\frac{X^2}{1+2X^2+X^4}$. Clearly this quotient also has a denominator that increases faster than its numerator. But then as $X$ increases (i.e. as $x$ decreases to $0$), $\frac{X^2}{1+2X^2+X^4}$ must get smaller. So then for $0\leq x\leq1$, $\frac{x}{(x+1)^2}\leq\frac{1}{4}$ and this exhausts all cases.

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Assume $x > 0$. Then we have $$2x^{1/2} \le x+1$$ Square each side:

$$4x \le x^2 + 2x + 1$$

Subtract $4x$ from each side:$$0 \le x^2 - 2x + 1$$
Factor: $$0 \le (x-1)^2$$ Since we know that any real number squared is non-negative, this is a true statement and the proof is done.

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It is equivalent to prove that

$$\frac {x+1}{\sqrt {x}}\geq 2$$

or

$$\sqrt {x}+\frac {1}{\sqrt {x}}\geq 2$$

or if $t=\sqrt {x} $, $$f (t)=t+\frac {1}{t}\geq 2.$$

now $f'(t)=1-\frac {1}{t^2} $.

its minimum is attained at $t=1$.

thus, $$f (t)\geq f (1)=2.$$

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