If $a+b+c+d+e=0$ so $180(a^6+b^6+c^6+d^6+e^6)\geq11(a^2+b^2+c^2+d^2+e^2)^3$

Question

Let $a$, $b$, $c$, $d$ and $e$ be real numbers such that $a+b+c+d+e=0$. Prove that: $$180(a^6+b^6+c^6+d^6+e^6)\geq11(a^2+b^2+c^2+d^2+e^2)^3$$ I tried Holder, uvw and more, but without success.

Answer 1

Let $x = (a,b,c,d,e)$.

By homogeneity, let $g(x) = a^2+b^2+c^2+d^2+e^2 = 1$. Further, as stated in the question, $h(x) =a+b+c+d+e=0$.

Let $$f(x) = a^6+b^6+c^6+d^6+e^6 $$ and we have to show, under the conditions above, that $f(x) \geq 11/180$. For $x_0 = \frac{1}{\sqrt{30}}(-3,-3,2,2,2)$ we have equality, $f(x_0)=11/180$, and also for all permutations in $x_0$, and the inversion to $-x_0$. These are $2 \cdot \binom{5}{2}$ = 20 points with $f=11/180$.

The proof is geometric. Consider the body $f(x) = $const. . This is the hypersurface of what has been called an N(5)-dimensional ball in p(6)-norm, see here. A nice visualization is given in here. Notice that an N-dimensional ball in p($\infty$)-norm is a hypercube. These balls are convex bodies, hence they have extremals where for given $f(x) = $const. , the Euclidean distance from the origin $g(x)$ has a maximum. Clearly, without further restrictions, these extrema occur at $x_1 = \frac{1}{\sqrt{5}}(\pm 1,\pm 1,\pm 1,\pm 1,\pm 1)$.

Due to the second condition $h(x)$, we need to consider the normal projection of these extrema $x_1$ to the space normal to $\frac{1}{\sqrt{5}}(1,1,1,1,1)$. An illustration can be given for the three-dimensional case, i.e. $f_3(x) = a^6+b^6+c^6 =$const. under the condition $h_3(x) =a+b+c=0$. We have $g_3(x) = a^2+b^2+c^2$. The normal space here is a plane, and the projections of $g_3$ (unit circle, green) and $f_3$ (red) onto this plane are given in the following figure, with a suitably chosen constant:

With a suitably chosen constant, $g_3$ and $f_3$ will touch at 6 positions, since there are 6 equivalent extrema (out of the 8 extrema $x_1$, 2 are parallel to $(1,1,1)$ and the other 6 are equivalent in value).

Back to the problem, for $x_{10} = \frac{1}{\sqrt{5}}(1,1,1,1,1)$, the normal projection is the nullvector. Considering one minus-sign, we have e.g. $x_{11} =\frac{1}{\sqrt{5}}(-1,1,1,1,1)$ and the projection is the vector $n_{11} = \frac{1}{\sqrt{20}}(-4,1,1,1,1)$. Considering two minus-signs, we have e.g. $x_{12} =\frac{1}{\sqrt{5}}(-1,-1,1,1,1)$ and the projection is the vector $n_{12} = \frac{1}{\sqrt{30}}(-3,-3,2,2,2)$. Further cases are given by permutations and inversions. The other cases of 3,4, and 5 minus-signs follow by inversion to $-x_{1n}$.

It is easy to see that indeed, for $f(x)=$const. , the value for the euclidian distance $g(x)$ for $n_{12}$ is the largest one for the $n$-vectors and therefore, by convexity, the largest value that $g(x)$ attains at all. Since we know already that $f(x) = 11/180$ holds with equality at that point (rather at the 20 equivalent points), and that the Hessian at this point is positive definite, this proves the claim. $\qquad \Box$