What is the minimum value for this equation?

The problem can be solved using AM-GM inequality as follows:

$\begin{aligned} A & = \frac {x+5}{\sqrt x+2} \\ & = \frac {(\sqrt x + 2)^2 - 4\sqrt x + 1}{\sqrt x + 2} \\ & = \frac {(\sqrt x + 2)^2 - 4(\sqrt x + 2) + 9}{\sqrt x + 2} \\ & = \blue{\sqrt x + 2} - 4 + \blue{\frac 9{\sqrt x +2}} & \small \blue{\text{By AM-GM inequality}} \\ & \ge \blue{2\sqrt{(\sqrt x+2)\cdot \frac 9{\sqrt x+2}}} - 4 = 6 - 4 = \boxed 2 \end{aligned}$

Equality occurs when $x = 1$ .

Given that $\displaystyle A(x) = \frac{x+5}{\sqrt{x}+2}$ , we have its domain $D_{A} = \{x \in \R : x \geq 0\}$ .

$\displaystyle \frac{dA(x)}{dx} = \frac{x + 4 \sqrt{x} - 5}{2 (\sqrt{x} + 2)^2 \sqrt{x}}$

$\displaystyle \frac{d^2A(x)}{dx^2} = \frac{10 + 15 \sqrt{x} - 6 x - x^{\frac{3}{2}}}{4 (2 + \sqrt{x})^3 x^{\frac{3}{2}}}$

With $\displaystyle D_{\frac{dA(x)}{dx}} = D_{\frac{d^2A(x)}{dx^2}} = (0 ; + \infty)$

$\displaystyle A_{min} \iff \begin{cases} \frac{dA(x)}{dx} =0 \\ \frac{d^2A(x)}{dx^2} > 0 \end{cases} \iff x = 1 \implies \orange{A_{min} = \boxed{2}}$ .

Note: I use the Calculus method to solve this Algebra problem.

Note that $A$ is real only when $x \geq 0$ .

$A=\frac{x+5}{\sqrt{x}+2}=\frac{(x-4)+9}{\sqrt{x}+2} =\frac{(\sqrt{x}-2)(\sqrt{x}+2)+9}{\sqrt{x}+2}$

$A=\sqrt{x}-2 +\frac{9}{\sqrt{x}+2} = [(\sqrt{x}+2)+\frac{9}{\sqrt{x}+2}] - 4$

Using the AM-GM inequality, we have:

$A \geq 2 \sqrt{(\sqrt{x}+2).\frac{9}{\sqrt{x}+2}}-4 =2\sqrt{9}-4 = 2$

As $A \geq 2$ , the minimum value of $A$ is $\boxed{2}$

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The equality happens when:

$\sqrt{x}+2 = \frac{9}{\sqrt{x}+2} \Leftrightarrow (\sqrt{x}+2)^2=9 \Leftrightarrow \sqrt{x}+2 = 3 \Leftrightarrow \sqrt{x} = 1 \Leftrightarrow x=1$

Since we have a square root sign, $x \geqslant 0$ . So we can let $x = X^2$ for non-negative $X$ . The expression becomes $A = \dfrac{X^2 + 5}{X + 2} = \dfrac{X^2 - 4 + 9}{X+2} = (X-2) + \dfrac9{X+2} = (X+2) + \dfrac9{X+2} - 4 .$ Using AM-GM inequality , $A + 4 \geqslant 2 \sqrt{ (X+2) \cdot \dfrac9{X+2} } = 6 \quad \implies A \geq\boxed{2}.$ The minimum occurs when $X + 2 = \frac9{X+2} \Leftrightarrow x = 1$ .