# Find the supremum of $A=(a^{2012}+b^{2012}+c^{2012}\mid a+b+c=1; a, b, c>0)$

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!

-
I’d really like to know why this was downvoted. – Brian M. Scott Nov 8 '12 at 21:21
@BrianM.Scott: Does seem weird, but then there are plenty of weird downvotes (up too). In this case, the post showed clear evidence of useful work towards a solution, so only upvoting makes sense. – André Nicolas Nov 8 '12 at 22:51
Is it possible to evaluate it somehow with means, so that we get solution? – fdhd Nov 8 '12 at 23:15

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 $1$.
Why you define $f(t)$ in such a way? Is it well known trick? If not, how can I guess the right way of defining similiar helpful functions? Beg your pardon, but I really want to understand it.. – fdhd Nov 8 '12 at 21:39
To 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. – Did Nov 8 '12 at 21:47
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 $1$.
Yes, 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. – fdhd Nov 8 '12 at 22:47