GM and AM convex cone

Let $$\displaystyle K_\alpha = \{x \in \mathbb{R}^n| \sqrt[n]{x_1x_2...x_n} \ge \alpha\frac{x_1 + x_2 + ... + x_n}{n}, x_i \ge 0\}$$ and $$\alpha \in [0, 1]$$. Show that $$K_\alpha$$ is a convex cone.

I tried just to use definition of convex cone and to check that $$\displaystyle\sqrt[n]{(\lambda_xx_1+\lambda_yy_1)(\lambda_xx_2+\lambda_yy_2)...(\lambda_xx_n+\lambda_yy_n)} \ge \alpha \frac{\lambda_x(x_1+x_2+...+x_n)+\lambda_y(y_1+y_2+...+y_n)}{n}$$ if $$\lambda_x\ge0, \lambda_y\ge0$$, but then I am stuck. How can this inequality be proved? Or should I use another approach?

Not too different from NN2's solution but I think this way is easier:

First we show that the set is a cone, then we show that it is convex.

To show that it's a cone we'll show that it is closed under positive scaling. Assume that $$a \in K_\alpha$$:

$$\left(\prod_i a_i\right)^{1/n} \geq \alpha \frac{\sum_i a_i}{n}$$

Then consider $$\lambda a$$ for $$\lambda > 0$$:

$$\left(\prod_i \lambda a_i\right)^{1/n} = \lambda\left(\prod_i a_i\right)^{1/n} \geq \lambda\alpha \frac{\sum_i a_i}{n} = \alpha \frac{\sum_i \lambda a_i}{n}$$

Next, to show that the set is convex, consider two elements of $$K_\alpha$$: $$x, y$$ and $$\lambda \in [0, 1]$$. Take the log of the LHS:

$$\frac{1}{n} \sum_i \log \lambda x_i + (1-\lambda)y_i \geq \lambda \frac{1}{n}\sum_i \log x_i + (1-\lambda) \frac{1}{n}\sum_i \log y_i$$

by Jensen's inequality. Now since both $$x, y \in K_\alpha$$ and exponentiating:

$$\left (\prod_ix_i^{1/n}\right )^\lambda \left (\prod_i y_i^{1/n}\right)^{(1-\lambda)} \geq \left (\alpha\frac{1}{n}\sum_ix_i\right )^\lambda \left (\alpha\frac{1}{n}\sum_i y_i\right)^{(1-\lambda)}$$

Finally, by the generalized AM-GM inequality (or by Jensen's inequality), we have that:

$$\geq \alpha\frac{1}{n}\sum_i\lambda x_i + (1-\lambda) y_i$$

So that the convex combination is also in $$K_\alpha$$.

The problem is solved if we succeed in proving

$$\frac{1}{n}\sum_{i=1}^n \ln\left(\lambda_x x_i+ \lambda_y y_i \right) \ge \ln\left( \lambda_x\sqrt[n]{\prod_{i=1}^nx_i} +\lambda_y\sqrt[n]{\prod_{i=1}^ny_i}\right) \tag{1}$$ (Indeed, the RHS of $$(1)$$ is greater than $$\ln\left( \lambda_x \frac{\alpha}{n} \sum_{i=1}^n x_i +\lambda_y\frac{\alpha}{n} \sum_{i=1}^n y_i\right)$$ )

We prove $$(1)$$ by using the Multivariate Jensen’s Inequality. This Jensen's inequality for two variables states that: if the function $$f:\matrix{\mathbb{R}^2 \mapsto\mathbb{R}\\(z,t)\mapsto f(z,t)}$$ is convex (the hessian matrix is positive semi definite, in other words, all the eigenvalues of the hessian matrix are nonnegative), then $$\frac{1}{n}\sum_{i=1}^nf\pmatrix{z_i\\t_i} \ge f\pmatrix{\frac{1}{n}\sum_{i=1}^nz_i\\\frac{1}{n}\sum_{i=1}^nt_i} \tag{2}$$

The inequality $$(1)$$ is equivalent to $$(1) \iff \frac{1}{n}\sum_{i=1}^n \ln\left(\lambda_x e^{\ln (x_i)}+ \lambda_y e^{\ln (y_i)} \right) \ge \ln\left( \lambda_x e^{\frac{1}{n}\sum_{i=1}^n\ln(x_i)} +\lambda_ye^{\frac{1}{n}\sum_{i=1}^n\ln(y_i)}\right) \tag{3}$$

We apply the Multivariate Jensen’s Inequality with the function $$f$$ defined as follows

$$f(z,t) := \ln\left(\lambda_x\cdot e^{z} + \lambda_y\cdot e^{t}\right)$$

This function is well convex as the hessian matrix has two nonnegative eigenvalues $$\frac{2\lambda_x \lambda_y e^{z+t}}{(\lambda_x e^z +\lambda_y e^t)^2}$$ and $$0$$.

Applying the result $$(2)$$ with $$(z_i,t_i) := \left(\ln(x_i), \ln(y_i) \right)$$ for $$i=1,..,n$$, we deduce that $$(3)$$ holds true. Then $$(1)$$ does also hold true.

Q.E.D

• Are you restricting the domain to be the positive orthant? Commented Sep 29, 2023 at 22:50
• @copper.hat Yes, the domain is well the positive orthant (as written by the OP in the question) $$\displaystyle K_\alpha = \{x \in \mathbb{R}^n| \sqrt[n]{x_1x_2...x_n} \ge \alpha\frac{x_1 + x_2 + ... + x_n}{n}, \color{red}{ x_i \ge 0}\}$$
– NN2
Commented Sep 29, 2023 at 23:00
• Thanks, I need to work on my reading skills :-) Commented Sep 29, 2023 at 23:09
• The function $x \mapsto \sqrt[n]{x_1...x_n}$ is concave on the positive orthant, so the map $f(x)= {\alpha \over n} (x_1+...+x_n) - \sqrt[n]{x_1...x_n}$ is convex. Hence the level set $f(x) \le 0$ is convex. Since $f$ is positively homogenous of degree one, it follows that the set is a convex cone. Commented Sep 29, 2023 at 23:20