my task is to find the supremum of the following set$A=(a^{2012}+b^{2012}+c^{2012}\mid a+b+c=1; a, b, c>0)$ I tried to use means method and it works for infimum (power mean vs. arithmetic mean) but I do not know which mean is greater than power (even used google...). I tried to play with $a=1, b=0, c=0$ and get supremum equals to $1$ but that is just guessing and might be wrong, so I ask you for any hints. Thanks in advance!
Find the supremum of A=(a^{2012}+b^{2012}+c^{2012}\mid a+b+c=1; a, b, c>0)
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0Is it possible to evaluate it somehow with means, so that we get solution? – 2012-11-08
2 Answers
Let $n=2012$, hence $n\gt1$. Let us fix $c$ in $[0,1]$ and let us maximize $a^n+b^n$ subject to $a+b=1-c$. One can define $f(t)=(t(1-c))^n+((1-t)(1-c))^n$ for every $t$ in $[0,1]$, then $f'(t)=n(t^{n-1}-(1-t)^{n-1})(1-c)^n$ has the sign of $t^{n-1}-(1-t)^{n-1}$, thus is negative if $t\lt\frac12$ and positive if $t\gt\frac12$. This shows that $f(t)$ is maximum at $t=0$ and/or at $t=1$. Both yield $f(t)=(1-c)^n$ hence one must now maximize $c^n+(1-c)^n$ on $c$ in $[0,1]$. The same reasoning yields the result that the maximum is achieved at $c=0$ and/or $c=1$. Finally, the maximum is $1$.
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1To maximize $a^n+b^n$ when $(a,b)$ are subject to the condition $a+b=1-c$, one can first *parametrize* such $(a,b)$, for example by $a=t(1-c)$ and $b=(1-t)(1-c)$ for $t$ in $[0,1]$. Then one is reduced to a function of a single variable, which are easier to maximize. – 2012-11-08
Note that $x^{2012}\le x$ on $[0,1]$, with equality only at $0$ and $1$. Since $a$, $b$, and $c$ are positive, $a^{2012}+b^{2012}+c^{2012}\lt a+b+c=1.$ Thus our sup is $\le 1$. But it is clear that by making $a$ very close to $1$, we can make $a^{2012}$ arbitrarily close to $1$.
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0Yes, I suspected that, but if I have (or guessed) the solution, I am obliged to prove it somehow, most likely with $\epsilon$ definition. I totally agree with your answer (thanks) but if it was a task in the exam, it would be not good. – 2012-11-08