Math Problem from a magazine

Method 1: By Titu's lemma

$\begin{aligned} \frac {1^2}a + \frac {1^2}a + \frac {1^2}a + \frac {1^2}a + \frac {1^2}{4b} & \ge \frac {(1+1+1+1+1)^2}{4(a+b)} = \frac {5^2}{4 \times \frac 54} = \boxed 5 & \small \color{#3D99F6} \text{Equality occurs when }a=1, b=\frac 14 \end{aligned}$

Method 2: By AM-HM inequality

$\begin{aligned} \frac 5{\frac 1a + \frac 1a + \frac 1a + \frac 1a + \frac 1{4b}} & \le \frac {a+a+a+a+4b}5 \\ \implies \frac 4a + \frac 1{4b} & \ge \frac {5^2}{4(a+b)} = \boxed 5 & \small \color{#3D99F6} \text{Equality occurs when }a=1, b=\frac 14 \end{aligned}$

Method 3: By Hölder's inequality

$\begin{aligned} \left(\frac 1a + \frac 1a + \frac 1a + \frac 1a + \frac 1{4b}\right)(a+a+a+a+4b) & \ge (1+1+1+1+1)^2 \\ \left(\frac 4a + \frac 1{4b}\right)(4)(a+b) & \ge 25 \\ \implies \frac 4a + \frac 1{4b} & \ge \boxed 5 & \small \color{#3D99F6} \text{Equality occurs when }a=1, b=\frac 14 \end{aligned}$

Method 1: Using Derivatives

$a + b = \frac{5}{4} \rightarrow b = \frac{5}{4} - a$

$P = \frac{4}{a} + \frac{1}{4(\frac{5}{4} - a)} = \frac{4}{a} + \frac{1}{5 - 4a}$

$P' = -\frac{4}{a^2} + \frac{4}{(5 - 4a)^2} = 0 \text{ (and } P'' > 0 \text{ for } 0 < a \leq \frac{5}{4}) \rightarrow a = 1 \rightarrow P = \boxed{5}$

Method 2: Using Derivatives

$a + b = \frac{5}{4} \rightarrow a = \frac{5}{4} - b$

$P = \frac{4}{\frac{5}{4} - b} + \frac{1}{4b} = \frac{16}{5 - 4b} + \frac{1}{4b}$

$P' = \frac{64}{(5 - 4b)^2} - \frac{1}{4b^2} = 0 \text{ (and } P'' > 0 \text{ for } 0 < b \leq \frac{5}{4}) \rightarrow b = \frac{1}{4} \rightarrow P = \boxed{5}$

Method 3: Using Graphing Technology

Graph $P = \frac{4}{a} + \frac{1}{5 - 4a}$ from Method 1

Method 4: Using Graphing Technology

Graph $P = \frac{16}{5 - 4b} + \frac{1}{4b}$ from Method 2

$P = \dfrac{4}{a}+\dfrac{1}{4b} \\ = \dfrac{4}{5}(a+b)(\dfrac{4}{a}+\dfrac{1}{4b}) \\ = \dfrac{4}{5}(4+\dfrac{1}{4}+\dfrac{4b}{a}+\dfrac{a}{4b}) \\ ≥ \dfrac{4}{5}(4+\dfrac{1}{4}+2\sqrt{\dfrac{4b}{a}·\dfrac{a}{4b}}) \ \ (By \ AM-GM \ Inequality) \\ = \dfrac{4}{5}×\dfrac{25}{4} \\ = \boxed{5}$

• Solution 1: 4/a + 1/4b = 4^2/4a + 1^2/4b >= (4+1)^2 / 4(a+b) = 5

• Solution 2: 4/a + 4a >= 8, 1/4b + 4b >= 2 => 4/a + 1/4b + 4(a+b) >= 10 => P >= 10 - 4*5/4 = 5

• Min P = 5 when a = 1, b = 1/4