Let $a,b,c$ be positive real numbers such that $a+b+c=2005$. Find the minimum value of the expression: $$E=a^{2006}+b^{2006}+c^{2006}+\frac{(ab)^{2004}+(bc)^{2004}+(ca)^{2004}}{(abc)^{2004}}$$
Problem
Source: Moldova national olympiad 2006, grade 9, pr.1
Tags: inequalities
dogofmath
21.09.2016 09:56
$E= a^2006 + \frac{1}{a^2004} + b^2006 + \frac{1}{b^2004} + c^2006 + \frac{1}{c^2004}$. By AM-GM $a^2006 + \frac{1}{a^2004} \ge 2a$ so $E\ge 2a+ 2b+ 2c = 4010$. The answer is 4010.
Snakes
21.09.2016 09:59
dogofmath wrote:
$E= a^2006 + \frac{1}{a^2004} + b^2006 + \frac{1}{b^2004} + c^2006 + \frac{1}{c^2004}$. By AM-GM $a^2006 + \frac{1}{a^2004} \ge 2a$ so $E\ge 2a+ 2b+ 2c = 4010$. The answer is 4010.
Almost all students at the olympiad did this mistake: In your solution the equality case is when $a=b=c=1$ but in this case we won't have $a+b+c=2005$...
sqing
21.09.2016 12:09
Snakes wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=2005$. Find the minimum value of the expression: $$E=a^{2006}+b^{2006}+c^{2006}+\frac{(ab)^{2004}+(bc)^{2004}+(ca)^{2004}}{(abc)^{2004}}$$ quykhtn-qa1 have: $ (ab)^{2004}+(bc)^{2004}+(ca)^{2004} \ge 3 \sqrt[3]{(abc)^{4008}}$ and $ abc \le \frac{2005^3}{3^3}$, $ \sum \ a^{2006}\ge \frac{2005^{2006}}{9}$ Then$ LHS \ge \frac{2005^{2006}}{9}+ \frac{3^{2005}}{2005^{2004}}$
rmtf1111
21.09.2016 18:53
Radon does all the job here: $(\star)\sum{a^{2006}}\ge \frac{2005^{2006}}{3^{2005}}$ $(\star\star)\sum{\frac{1}{a^{2004}}}\ge \frac{3^{2005}}{2005^{2004}}$ Summing $(\star)$ and $(\star\star)$ we get the minimum value of $E$.
rmtf1111
21.09.2016 18:55
sqing wrote: $ \sum \ a^{2006}\ge \frac{2005^{2006}}{9}$ Isn't this wrong?
weirdo37
21.09.2016 20:54
Its just Jensen.