AM-GM inequality

Algebra Level 5

$3\sqrt{3}a^3+\frac{2}{ab-2b^2}$

Find the minimum value of the above expression if $a>2b>0$

The answer is 20.

1 solution

Chan Lye Lee
Mar 31, 2016

Note that $\frac{2b+(a-2b)}{2} \ge \sqrt{2b(a-2b)}$ by AM-GM. This leads $\frac{1}{b(a-2b)}\ge\frac{8}{a^2}$ . Now $3\sqrt{3}a^3+\frac{2}{ab-2b^2} \ge 3\sqrt{3}a^3+\frac{16}{a^2} = \frac{3\sqrt{3}a^3}{2}+\frac{3\sqrt{3}a^3}{2}+\frac{16}{3a^2}+\frac{16}{3a^2}+\frac{16}{3a^2}$

$\ge 5 \left( \frac{3\sqrt{3}a^3}{2} \times \frac{3\sqrt{3}a^3}{2} \times \frac{16}{3a^2} \times \frac{16}{3a^2}\times \frac{16}{3a^2} \right)^\frac{1}{5} = 20$

Nice solution. Equality holds when $\dfrac{3\sqrt{3}a^{3}}{2} = \dfrac{16}{a^{2}} \Longrightarrow a = \dfrac{2}{3}\sqrt{3}$ and $b = \dfrac{\sqrt{3}}{6}$ , ( $a = 4b$ ).

Brian Charlesworth - 5 years, 2 months ago

What's the problem in this?

Let $a-2b=x$ .

Then we have the above expression as

$3\sqrt{3} ( 8b^{3} + x^{3} + 12b^{2} x + 6bx^{2}) + 1/3bx + 1/3bx + 1/3bx + 1/3bx + 1/3bx + 1/3bx$

Now apply $AM\geq GM$ .

Then we have its minimum value as

$10(576)^{1/10}$ = $18.888 >18$ .

Aakash Khandelwal - 5 years, 2 months ago

Usually it is because the equality never hold . it is nothing wrong to say it is >18.888, but the problem is we can not get the value (for any a and b).

Chan Lye Lee - 5 years, 2 months ago

AM-GM inequality

The answer is 20.

1 solution

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