Prove x13+x23+x33+x43 ≥ x1+x2+x3+x4 with x1x2x3x4=1

[asmani]

Suppose that x₁,x₂,x₃,x₄ are four positive real numbers such that x₁x₂x₃x₄=1.

Show that x₁³+x₂³+x₃³+x₄³ ≥ x₁+x₂+x₃+x₄.

How to solve this? (Not a homework!)

Any hint is appreciated.

 

[JonF]

Looks like it should be pretty easy with induction if you can prove the base case of:

if x₁,x₂ are positive real numbers such that x₁x₂=1 then x₁³ + x₂³≥ x₁ + x₂

For the base case you know that x₁x₂=1 ⇒ x₂= 1/x₁

i.e. show x³ + 1/x³ - x - 1/x ≥ 0, for x ≥ 0 which shouldn't be to bad with some calculus

At least that's the route I would go.

 

[dalcde]

MO Question?

 

[disregardthat]

Use lagrange. It will pop right out.

 

[asmani]

Thanks for the replies.

That's a part of a problem of a national mathematical olympiad. The full problem is as follows:

Suppose that x₁,x₂,x₃,x₄ are four positive real numbers such that x₁x₂x₃x₄=1. Show that

 

[Anonymous217]

Looks like it should be pretty easy with induction if you can prove the base case of:

if x₁,x₂ are positive real numbers such that x₁x₂=1 then x₁³ + x₂³≥ x₁ + x₂

For the base case you know that x₁x₂=1 ⇒ x₂= 1/x₁

i.e. show x³ + 1/x³ - x - 1/x ≥ 0, for x ≥ 0 which shouldn't be to bad with some calculus

At least that's the route I would go.

Base case could just be n=1, i.e. one x. I guess you could also argue for n=0 too.

 

[JonF]

Base case could just be n=1, i.e. one x. I guess you could also argue for n=0 too.

good point, that makes it even easier

 

[Dr. Seafood]

Base case could just be n=1, i.e. one x. I guess you could also argue for n=0 too.

Well, you got to watch out for which natural number corresponds to the base case when carrying out inductive proofs.

Check this out: We will show n^3 \leq n^2 for all n > 0 by induction. If n = 1, we get 1^3 \leq 1^2 which is certainly true. Now assume k^3 \leq k^2. Then

(k + 1)^3 = k^3 + 3k^2 + 3k + 1 \leq k^2 + 3k^2 + 3k + 1 \leq k^2 + 2k + 1 = (k + 1)^2

which was the desired result. But wait, 2^3 = 8 > 4 = 2^2. Contradiction? The problem comes from "choice" of base case.

 

[disregardthat]

Analytical solution:

Let f(x_1,x_2,x_3,x_4) = x_1^3+x_2^3+x_3^3+x_4^3-(x_1+x_2+x_3+x_4), and g(x_1,x_2,x_3,x_4) = x_1x_2x_3x_4-1.

we will find the minimum of f under the condition g = 0.

We find the solutions to the lagrange equations

\frac{\delta f}{\delta x_i} = \lambda \frac{\delta g}{\delta x_i}

and g = 0, or

3x_i^2 -1= \lambda x_1x_2x_3 \Leftrightarrow 3x_i^3-x_i=\lambda, so 3x_i^3-x_i = 3x_j^3-x_j, or (x_i-x_j)(3(x_i^2+x_ix_j+x_j^2)-1) = 0. Pick the largest one, say x_k. It will be larger or equal than 1. And thus 3(x_k^2+x_ix_k+x_k^2)-1 \geq 2 > 0. Hence x_i = x_k for all i, so they are all equal. Hence x_i = 1 for all i.

Now this is the minimum on some compact set containing (1,1,1,1).

The set determined by x_i > 0, and x_1x_2x_3x_4 = 1, and x_1+x_2+x_3+x_4 <=5 contains (1,1,1,1) and is compact.

Furthermore, by the general QM-AM \sqrt[3]{\frac{x_1^3+x_2^3+x_3^3+x_4^3}{4}} \geq \frac{x_1+x_2+x_3+x_4}{4}

So x_1^3+x_2^3+x_3^3+x_4^3 \geq \frac{(x_1+x_2+x_3+x_4)^2}{16} (x_1+x_2+x_3+x_4)

Hence outside our set f(x_1,x_2,x_3,x_4) > 0. So (1,1,1,1) is actually the value for which f is minimized, and the value is 0. Hence x_1^3+x_2^3+x_2^3+x_4^3 \geq x_1+x_2+x_3+x_4.

(EDIT: I assumed we were working in the subset x_i > 0 originally, so just assume I said that before I did :) )

 

[disregardthat]

Well, you got to watch out for which natural number corresponds to the base case when carrying out inductive proofs.

Check this out: We will show n^3 \leq n^2 for all n > 0 by induction. If n = 1, we get 1^3 \leq 1^2 which is certainly true. Now assume k^3 \leq k^2. Then

(k + 1)^3 = k^3 + 3k^2 + 3k + 1 \leq k^2 + 3k^2 + 3k + 1 \leq k^2 + 2k + 1 = (k + 1)^2

which was the desired result. But wait, 2^3 = 8 > 4 = 2^2. Contradiction? The problem comes from "choice" of base case.

k^2 + 3k^2 + 3k + 1 \leq k^2 + 2k + 1 that does follow from the induction hypothesis and is certainly not true.

 

[Dr. Seafood]

That's what I said.

 

[disregardthat]

That's what I said.

No, you derived those inequalities in your post, but it's wrong. How does it have anything to do with the base case?