Let Our Powers Combine

Alternatively, we want to find the smallest n such that there exist some $x_i$ with $\sum_{i=1}^n x_i^6=n$ and $\sum_{i=1}^n x_i^5-\sum_{i=1}^n x_i^2 <0$ . So, first, I will find how a minimum solution for any $n$ looks like, then I will solve the $n$ that gets that minimum to less than 0.

Let $f(t)=t^{5/6}-t^{1/3}$ (we only use t>0) then we want to minimize $\sum_{i=1}^n f(y_i)$ , with $y_i=x_i^6$ and $\sum_{i=1}^n y_i=n$ .

We will try to use Jensen (Or, rather, Karamata (http://en.wikipedia.org/wiki/Karamata's_inequality)). We should note that $f''(t)=\frac{8-5 t^{1/2}}{36 t^{5/3}}$ and that is positive if $0<t<\frac{64}{25}$ and negative when $\frac{64}{25}<t$ .

Karamata inequality, roughly, says that if $f$ is convex then $\sum_{i=1}^n f(\alpha_i)$ . given that the sum of the $\alpha$ 's is constant, when all the $\alpha_i$ are equal. When is concave, it minimizes when they are as further apart as they can. So let the solution that minimizes it be $\{y_i\}=\{a_i\} \bigcup\{b_i\}$ , with the a's less than 64/25, and the $b$ 's greater than. Note that we can consider 64/25 's in any of the sets we want, but we will prefer to see them as $a$ 's.

Because, in the domain $(0,64/25]$ $f$ is convex, then all the a's must be equal because if not, Karamata inequality would find a solution with lower sum of $f$ .

Because in the domain $[64/25,\infty)$ $f$ is concave, then if there are at least 2 $b$ 's, one must be at the extreme, otherwise Karamata will give us a lower solution. But then we can consider it rather an a and not an. So there it is at most one $b$ .

So the minimal solution is $(a,a,..a,b)$ with $n-1$ $a$ 's and one $b$ , with $a$ and $b$ not necessarily different. If they are different, though, then $b>64/25>a$ . So overall $b\geq a$

Now we will proceed to find the $n$ that works. For simplicity, I will suppose there are n+1 terms (So that I can use n a's and one b).

We will use the technique of homogenization(http://www.artofproblemsolving.com/Wiki/index.php/Homogenization). Because $x_i>0$ then the inequality is equivalent to squaring it and multiply it by (n+1) so it is equivalent to $(n+1)(\sum_{i=1}^{(n+1)} x_i^5)^2\geq (n+1) (\sum_{i=1}^{(n+1)} x_i^2)^2$ . Using the hypothesis on the right side, then $(n+1)(\sum_{i=1}^{(n+1)} x_i^5)^2\geq (\sum_{i=1}^{(n+1)} x_i^6) (\sum_{i=1}^{(n+1)} x_i^2)^2$ . Now the inequality is homogeneous, that is, every term is of the same degree. So if we multiply every number by a positive constant $k$ then both sides are multiplied by a positive constant $C_k$ , and it is equivalent so we can forget the hypothesis.

Remembering what we prove, it is true for every {x_i} if and only if is true for every $(a,a,...,a,b)$ , for any $a,b$ . So we will plug it in the inequality. So the inequality is equivalent to $(na^2+b^2)^2(na^6+b^6)\leq (na^5+b^5)(n+1)$ . If a=0 it is trivially true. Because homogeneity, we can make an assumption and retain generality. So we will assume $a=1$ and now we know that $b\geq a=1$ (we can because a is not 0) and so the inequality is equivalent to $(n+b^2)^2(n+b^6)\leq (n+b^5)(n+1)$ . We pass the left side to the right side and factor it to $(b-1)^2(b^2+b+1)(b^6+b^5-b^4-n(b^2-b-1))$ . We know $(b-1)^2,b^2+b+1$ are always positive, so the whole inequality is equivalent to $b^6+b^5-b^4-n(b^2-b-1)\geq 0$ . If $b^2-b-1$ is not positive, then it is also trivially true because it would be positive plus positive. So we can asume is $b^2-b-1\geq 0$ then $b>\phi$ .

So the inequality is, finally, equivalent to $n\leq \frac{b^6+b^5-b^4}{b^2-b-1}=g(b)$ for any $b> \phi$ . Because $g$ it is well defined in $(\phi,\infty)$ , has no singularities and the extremes tend to positive infinity, then the global minimum is also a local minimum. So we have to find the minimum of $g$ . With a little of calculus and factorizing it, we see that $g'(t)=0$ at $t=-1,0,1/2,2$ . Of those, only 2 is greater than $\phi$ (also $g''(2)>0$ ). So that would be the minimum , and at that point $g(2)=80$ . So we have that the inequality is true for $n+1$ variables if and only if $n\leq g(b)$ for any b in $(\phi, \infty)$ if and only if $n\leq \min g(b)$ if and only if $n\leq 80$ .

So the largest number of variables allowed is $80+1=81$

81