## Saturday, September 30, 2006

### Eww... Topic: Algebra/Inequalties. Level: AIME/Olympiad.

Problem: Find the minimum value of

$\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{ \left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}$

for $x > 0$.

Solution: Make a handy substitution $a = x$, $b = \frac{1}{x}$ and $1 = a^3b^3$. Then we want to minimize

$\frac{(a+b)^6-(a^6+b^6)-2a^3b^3}{(a+b)^3-(a^3+b^3)} = \frac{2a^4+5a^3b+6a^2b^2+5ab^3+2b^4}{a+b}$

after cancelling $3ab$. But we see that this evenly divides and becomes

$2a^3+3a^2b+3ab^2+2b^2 = (a+b)^3+(a^3+b^3)$.

But $a+b \ge 2$ and $a^3+b^3 \ge 2$ by AM-GM and the same equality case, so

$(a+b)^3+a^3+b^3 \ge 10$

with equality at $a = b = 1$. QED.

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Comment: It was very useful to make the substitution and then homogenize the numerator so you could cancel things out evenly. Then classical inequalities (AM-GM) helped us finish it.

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Practice Problem: Prove that $(a^3+b^3+c^3-3abc)(a+b+c) \ge 0$.

#### 2 comments:

1. I'm pretty sure you can do the first problem by substituting y = x + 1/x and then using y \ge 2 as your last step.

(a + b + c)^2( (a-b)^2 + (b - c)^2 + (c-a)^2 ) \ge 0 trivially.

2. Links

How much does it cost links on your blog (Blogroll)?