Arithmetic Mean-Geometric Mean inequality question

[YouAreAwesome]

Homework Statement

Textbook example states that (a/b + b/c + c/d + d/a)/4 ≥ (a/b × b/c × c/d × d/a)^(1/2) based on AM-GM. But AM-GM states that the RHS should be the 4th root i.e ^(1/4), not the square root, because there are 4 terms. I thought this was a typo, but then a later example uses the exact same idea.

Relevant Equations

For numbers x₁, x₂, ..., xₙ: (x₁ + x₂ + ... + xₙ) / n ≥ (x₁ * x₂ * ... * xₙ)^(1/n)

Hi,

Can someone explain why we can use the square root rather than the 4th root in this application of the AM-GM?

(a/b + b/c + c/d + d/a) / 4 ≥ (a/b × b/c × c/d × d/a)^(1/2)

Thanks

 

[FactChecker]

I think it is a typo. But it doesn't matter in this example because the RHS is 1.
Is the later example similar? If its RHS is $\le 1$, then the fourth root is $\ge$ the square root.

 

[YouAreAwesome]

I think it is a typo. But it doesn't matter in this example because the RHS is 1.
Is the later example similar?

Great thanks, I was satisfied it was a typo, or that the writer recognized the product was equal to 1 so thought it didn't matter. But then in a later proof he writes:

$$

\frac{\frac{a^3}{b} + \frac{b^3}{c} + \frac{c^3}{d} + \frac{d^3}{a}}{4} \geq \sqrt{\frac{a^3}{b} \times \frac{b^3}{c} \times \frac{c^3}{d} \times \frac{d^3}{a}}

$$
$$
\frac{\frac{a^3}{b} + \frac{b^3}{c} + \frac{c^3}{d} + \frac{d^3}{a}}{4} \geq \sqrt{a^2b^2c^2d^2}

$$
$$

\frac{a^3}{b} + \frac{b^3}{c} + \frac{c^3}{d} + \frac{d^3}{a} \geq 4abcd

$$

Note:

$$ a,b,c,d > 0 $$

 

[YouAreAwesome]

I think it is a typo. But it doesn't matter in this example because the RHS is 1.
Is the later example similar? If its RHS is $\le 1$, then the fourth root is $\ge$ the square root.

Ok, I think there are errors in the textbook. I just tried a simple substitution of $a=1, b=2, c=3, d=4$ and the conclusion is false. $4abcd = 96$ while the $LHS= 74$ approximately. So the RHS should have $\sqrt{abcd}$ and I've been lead down a rabbit hole.

Apologies. I wonder if I should delete this thread?

 

[berkeman]

Your thread is fine.