# Looking for a way to prove that the exponential mean of order alpha <= the geometric mean

pic of problem

I'm looking for a way to just get started on proving this inequality. I initially was thinking that I might be able to use Jensen's inequality to help prove this, but I'm at a loss as to how I would apply it. Any guidance in the right direction would be really helpful!

• Are you sure of the inequality sign. I am guessing that you can use AM-GM relationship but the sign goes the opposite way!! – Satish Ramanathan Dec 3 '17 at 23:33
• I am sure of the direction of the sign. However, as I mentioned in a comment below, x1 can be different value than xn, which nullifies the use of the AM-GM relationship. Also, alpha is a negative real number, which is going to be significant somehow – Peter Olson Dec 4 '17 at 0:33

## 1 Answer

Taking the power $\alpha$ on both sides of the equation gives $$\frac{(x_1)^{\alpha}+(x_2)^{\alpha}+\cdots+(x_n)^{\alpha}}{n}\leq \left(x_1 x_2 \cdots x_n \right)^{\frac{\alpha}{n}} = \sqrt[n]{(x_1)^{\alpha} (x_2)^{\alpha} \cdots (x_n)^{\alpha}}.$$ Considering the change $y_1=(x_1)^{\alpha},y_2=(x_2)^{\alpha},\cdots,y_n=(x_n)^{\alpha}$, it transforms the inequality to $$\frac{y_1+y_2+\cdots+y_n}{n} \leq \sqrt[n]{y_1 y_2 \cdots y_n}.$$ The left term is the arithmetic mean, the right the geometric mean. Therefore this inequalty doesn't hold, as the geometric mean is lesser or equal than the arithmetic mean. (See wikipedia on the inequality of arithmetic and geometric means).

• I had come across the AM-GM relationship, but I don't think it can be applied here since x1...xn need not be equal to one another. That is, the problem states that x1, x2, ..., xn are any fixed positive real numbers. Additionally, another part of the problem which I chose not to inquire about since it is quite similar, is to prove that the geometric mean is less than the exponential mean of order beta, where beta is a positive real number (noting that alpha is a negative real value) – Peter Olson Dec 4 '17 at 0:05