Maximizing the Expression with $a+b+c=2012$

[anemone]

Let $a,\,b$ and $c$ be positive real numbers where $a+b+c=2012$.

Find the maximum of \(\displaystyle \frac{(a^2+b^2+c^2)(a^3+b^3+c^3)}{a^4+b^4+c^4}\).

 

[kaliprasad]

Let $a,\,b$ and $c$ be positive real numbers where $a+b+c=2012$.

Find the maximum of \(\displaystyle \frac{(a^2+b^2+c^2)(a^3+b^3+c^3)}{a^4+b^4+c^4}\).

My solution

Spoiler

from symmetry it is maximum when $a=b=c$ and we get
$\frac{(a^2+b^2+c^2)(a^3+b^3+c^3)}{a^4+b^4+c^4}= \frac{3a^2 * 3a^3}{3a^4} = 3a = 2012$
this is maximmum because at $1,1,2010$ we get
$\frac{(a^2+b^2+c^2)(a^3+b^3+c^3)}{a^4+b^4+c^4} = \frac{(1 + 1 + 2010^2)(1 + 1 + 2010^3)}{1+1+ 2010^4} = 2010 < 2012$
hencer maximum = 2012 when $a=b=c = \frac{2012}{3}$

 

[anemone]

Good job kaliprasad!(Cool)