GM and AM convex cone

Question

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?

Answer 1

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

Answer 2

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 $K_\alpha$.