# Calculus inequality $\sqrt{x}\leqslant \frac14x+1$

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?

• $$f'(x)=\frac14-\frac1{2\sqrt x}\geq0,\,\forall x\geq4.$$ Commented Apr 2, 2020 at 13:30
• How do we come up with the condition for $x$ to be either greater than $4$ or $0$?
– user745970
Commented Apr 2, 2020 at 13:34
• 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 $x\ge0$. Commented Apr 2, 2020 at 13:37
• @WETutorialSchool for $\text{AM-GM}$ how do we get $n = 2$ since we have only one variable $x$?
– user745970
Commented Apr 2, 2020 at 14:10
• 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 $b=1$. Commented Apr 2, 2020 at 14:12

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 $$y \geq 2 \Leftrightarrow x \geq 4$$.

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 $$f(x)\le f(4)=0 \implies \sqrt{x}\le \frac{x}{4}+1.$$

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 $$x\geq0$$.

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.

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

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 $$4 < x$$.