# Proving that $(x+y+2z)(2x+4y+z)(x+y+z)$ defines a convex level set

How can I prove that the following function is concave over $x,y,z\ge 1$?$$(x+y+2z)(2x+4y+z)(x+y+z)$$

I tried the following: since $(\prod _{i=1}^n x_i)^ {\frac 1n}$ is concave over $\mathbb R^n_+$, then $$((x+y+2z)(2x+4y+z)(x+y+z))^{\frac 13}$$is also concave. But the function $t^3$ is convex over $t\ge0$ and not concave so I got stuck.

I also tried to to show that the hessian matrix is negative semidefinite, but it is just too tidious. Any ideas?

Edit

It was pointed out in the answers that the function is not concave. However, maybe it is concave over the feasible set defined by the intersection of the following level sets? $$(x+y+2z)(2x+4y+z)(x+y+z)\ge1$$ $$x^2+y^2+2z^2+2xy+2xz+2yz\le 1$$ $$e^{e^x}+\max\{0,y\}^3\le 7$$ $$x,y,z\ge 0.1$$

Clearly the second, third and fourth level sets are convex.
Is it possible the the first function is indeed concave over the set defined by the other constraints? (and thus defines a convex level set). I'm trying to figure out if there is something wrong with this question or whether I'm missing something trivial.
Maybe the first function is quasi-concave thus defines a convex level set?

• I would use the inequality of Jensen – Dr. Sonnhard Graubner Jul 7 '18 at 20:05
• – Dr. Sonnhard Graubner Jul 7 '18 at 20:09
• How would you define the function $f$? – MasterJ Jul 7 '18 at 20:39
• As the second answer points out: for the Hessian(in the region indicated), there are one positive eigenvalue and two negative eigenvalues. Your function is not convex, but it is not concave either. Calculate the Hessian matrix at the point $(1,1,1)$ for easy numbers, then find out the signs of the eigenvalues. – Will Jagy Jul 7 '18 at 23:40

It's not concave: On the line $(x,y,z) = (t,t,t)$ the function is equal to $84t^3$. The second derivative of $84t^3$ is $504t$, which is positive when $t > 0$. Hence your function is convex when restricted to this ray, and therefore cannot be concave overall in your domain since this domain intersects the ray in the set where $t \geq 1$.

• that's what I thought after fiddling with it for a time – Will Jagy Jul 7 '18 at 23:16
• Thanks. I've edited the question. Is it possible that function is not concave but defines a convex level set? – MasterJ Jul 8 '18 at 8:40

There is a theorem, special to this dimension. The Hessian matrix entries are linear functions, homogeneous degree one. You are given a cubic form that completely factors. Here is the theorem: the determinant $\sigma_3$ of the Hessian matrix, which is also a cubic form, is a constant multiple of the original form you were given. In the region specified, your form is always positive.

Next, the trace \sigma_1 of the Hessian is a linear form; the sum of the principal two by two matrix determinants $\sigma_2$ is a quadratic form. Finally, the characteristic polynomial of the Hessian is $\lambda^3 - \sigma_1 \lambda^2 + \sigma_2 \lambda - \sigma_3.$

• Thanks. Do you have a reference to this theorem? – MasterJ Jul 7 '18 at 20:37
• books.google.com/… – Will Jagy Jul 7 '18 at 20:41

The first constraint is a level set of the convex function $$-\log(x_1+x_2+2x_3)-\log(2x_2+4x_2+x_3)-\log(x_1+x_2+x_3)$$and the other constraints are convex.