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

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The discussion focuses on maximizing the expression \((a^2+b^2+c^2)(a^3+b^3+c^3)/(a^4+b^4+c^4)\) under the constraint \(a+b+c=2012\) with \(a, b, c\) as positive real numbers. Participants explore various mathematical strategies and inequalities to derive the maximum value of the expression. The conversation highlights the importance of symmetry and optimization techniques in solving the problem. The thread concludes with a positive acknowledgment of contributions made by participants. The mathematical exploration emphasizes the complexity and beauty of optimization in algebraic expressions.
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Let $a,\,b$ and $c$ be positive real numbers where $a+b+c=2012$.

Find the maximum of $$\frac{(a^2+b^2+c^2)(a^3+b^3+c^3)}{a^4+b^4+c^4}$$.
 
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anemone said:
Let $a,\,b$ and $c$ be positive real numbers where $a+b+c=2012$.

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

My solution
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 $a=b=c = \frac{2012}{3}$
 
Good job kaliprasad!(Cool)
 
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