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Show that for all $x \geqslant 4$ the following inequality holds $$\sqrt{x}\leqslant \frac14x+1.$$ Hint: $f(x) = \frac14x+1-\sqrt{x}$

So if we denote $f(x) = \frac14x+1-\sqrt{x}$.

We could just show that $f'(x) \geqslant0$ and that would satisfy the given inequality?

However $f'(x) = \frac14-\frac{1}{2\sqrt{x}}$ which doesn't hold when $x\geqslant 0$. What's the trick here?

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    $\begingroup$ $$f'(x)=\frac14-\frac1{2\sqrt x}\geq0,\,\forall x\geq4.$$ $\endgroup$
    – awllower
    Commented Apr 2, 2020 at 13:30
  • $\begingroup$ How do we come up with the condition for $x$ to be either greater than $4$ or $0$? $\endgroup$
    – user745970
    Commented Apr 2, 2020 at 13:34
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    $\begingroup$ This is just $\left(\frac{\sqrt{x}}{2}-1\right)^2\geq 0$. You can also simply use AM-GM: $\frac{x}{4}+1\geq 2\sqrt{\frac{x}{4}\cdot 1}=\sqrt{x}$. The inequality is true for all $x\ge0$. $\endgroup$ Commented Apr 2, 2020 at 13:37
  • $\begingroup$ @WETutorialSchool for $\text{AM-GM}$ how do we get $n = 2$ since we have only one variable $x$? $\endgroup$
    – user745970
    Commented Apr 2, 2020 at 14:10
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    $\begingroup$ Use AM-GM in the $2$-variable form: $\frac{a+b}{2}\ge \sqrt{ab}$. This is equivalent to $a+b\ge 2\sqrt{ab}$. Take $a=x/4$ and $b=1$. $\endgroup$ Commented Apr 2, 2020 at 14:12

5 Answers 5

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Just set $x=y^2$ and see

$$\sqrt{x} \leq \frac 14 x+ 1 \Leftrightarrow y\leq \frac 14 y^2 + 1$$ $$\Leftrightarrow 0\leq y^2-4y+4 = (y-2)^2$$

which is true for all $y \geq 2 \Leftrightarrow x \geq 4$.

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The calculus method

Let $$f(x)=\sqrt{x}-\frac{x}{4}-1 \implies f'(x)=\frac{1}{2\sqrt{x}}-\frac{1}{4}, ~~f''(x)=-\frac{1}{4}x^{-3/2}.$$ Then $$f'(x)=0 \implies x=4 \implies f''(4)<0$$ So $f(x)$ has only one local max, therefore $$f(x)\le f(4)=0 \implies \sqrt{x}\le \frac{x}{4}+1.$$

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When you are asked this type of inequality, you have to solve this system of inequality: $$\left\{\begin{matrix} x\geq0 \\ \frac{1}{4}x+1\geq0 \\ x^2-8x+16\geq0 \end{matrix}\right.$$ Solving, we have: $$\left\{\begin{matrix} x\geq0 \\ x\geq-4 \\ (x-4)^2\geq0 \end{matrix}\right.$$ Being $(x-4)^2$ always positive, we duduce that the solution is $x\geq0$.

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For $x\geq4$ we have $$\frac{1}{4}-\frac{1}{2\sqrt{x}}=\frac{\sqrt{x}-2}{4\sqrt{x}}\geq0,$$ which says that $f$ increases.

Thus, $$f(x)\geq f(4)=0,$$ which ends a proof.

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$$f(x) = \frac14x+1-\sqrt{x}$$

You get $f'(x) = \dfrac 14 - \dfrac{1}{2 \sqrt x}$ and $f''(x)=\dfrac{1}{4\sqrt{x^3}}$

So $f'(x)$

  • Is strictly increasing on the inverval $(0, \infty)$.
  • Has a vertical asymptote at $x=0$.
  • Passes through the point $(4,0)$.

So $f(x)$

  • Passes through the point $(0,1)$.
  • Is decreasing in the interval $0 \lt x \lt 4$.
  • Has a global minimum at $(4,0)$. (It follows that $f(x) \ge 0$.)
  • Is increasing on the interval $4 < x$.
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