Feb 27, 2014 · The arithmetic - geometric mean inequality states that x1 + … + xn n ≥ n√x1⋯xn I'm looking for some original proofs of this inequality. I can find the usual proofs on the internet but I was wondering if someone knew a proof that is unexpected in some way. e.g. can you link the theorem to some famous theorem, can you find a non-trivial geometric proof (I can find some …

Sep 1, 2020 · I recently came up with a proof by simple induction of the arithmetic mean - geometric mean inequality that I haven't found here. I'm sure it isn't new. My questions: (1) Is this correct? (2) Is th...

Sep 2, 2015 · In AM GM inequality for nonnegative real numbers a1,a2, …,an, How to show that if equality holds then a1 =a2 = … =an, using method of induction?

This question is from the textbook Introduction to Linear Algebra by Gilbert Strang, where the author has asked to prove that $$\\sqrt[3]{xyz} \\le \\frac{x+y+z}{3}. $$ The only equations at hand ar...

Jan 4, 2019 · I have two questions about this: Is this concept already known? I believe that the AM-GM inequality generalizes to ℓ1 ≥ℓ2 ≥ ⋯ ≥ℓn ℓ 1 ≥ ℓ 2 ≥ ⋯ ≥ ℓ n. Is this correct? Thank you!

Jun 10, 2015 · Hint: If you can use the AM-GM on 2 numbers, then you should use that. First prove with this inequality on (a, b) (a, b)

Jul 1, 2022 · To see this, first notice that weighted AM-GM works with fractional weights. This is because you can multiply each fraction by the lowest common denominator of all the fractions without changing either side of the inequality. Since you can approximate any real number arbitrarily close by fractions, weighted AM-GM works for real numbers too.

For questions about proving and manipulating the AM-GM inequality. To be used necessarily with the [inequality] tag.

Feb 19, 2020 · I was trying to find the minimum value of a + 1 a + 3 a + 1 a + 3 given that a a is a positive real number. I found 2 2 ways to do it. Minimum value of a + 1 a a + 1 a is 2 2 using the AM-GM inequality and therefore the minimum value of a + 1 a + 3 a + 1 a + 3 is 2 + 3 = 5 2 + 3 = 5 Then, I tried using AM-GM inequality directly for the given expression and I got a + 1 a + 3 ≥ …

Apr 20, 2016 · Many famous inequalities can be proved by the rearrangement inequality, such as the arithmetic mean – geometric mean inequality, the Cauchy–Schwarz inequality, and Chebyshev's sum inequality. How can the inequality of arithmetic and geometric mean be shown using rearrangement inequality? (I had a look at this post: Proofs of AM-GM inequality.