18+12+8+5 right?

Algebra Level 4

Find the minimum value of

$\large \frac{18}{a+b}+\frac{12}{ab}+8a+5b$

where $a$ and $b$ are positive real numbers .

The answer is 30.

1 solution

Chew-Seong Cheong
Oct 31, 2017

Rewrite the expression $\dfrac {18}{a+b} + \dfrac {12}{ab} + 8a+5b$ as follows and apply AM-GM inequality :

$\begin{aligned} {\color{#3D99F6}\frac {18}{a+b} + 2(a + b)} + {\color{#D61F06} \frac {12}{ab} + 6a+3b} \ge {\color{#3D99F6} 2\sqrt{\frac {18(2(a+b))}{a+b}}} + {\color{#D61F06} 3\sqrt[3]{\frac {12 (6a)(3b)}{ab}}} = {\color{#3D99F6}12} + {\color{#D61F06} 18} = \boxed{30} \end{aligned}$

Equality occurs when $a=1$ and $b=2$ .

How did you know that you should rewrite it like that and not, say, 18/(a+b) + 3(a+b) + 12/ab + 5a+2b?

Zainul Niaz - 3 years, 7 months ago

There is no rule. We have to try and error.

Chew-Seong Cheong - 3 years, 7 months ago

18+12+8+5 right?

The answer is 30.

1 solution

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