# An inequality

1. Aug 16, 2011

### 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.

2. Aug 16, 2011

### 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.

3. Aug 16, 2011

### dalcde

MO Question?

4. Aug 16, 2011

### disregardthat

Use lagrange. It will pop right out.

5. Aug 19, 2011

### 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

6. Aug 19, 2011

### Anonymous217

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

7. Aug 19, 2011

### JonF

good point, that makes it even easier

8. Aug 19, 2011

### Dr. Seafood

Well, you gotta 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.

9. Aug 19, 2011

### 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 :) )

Last edited: Aug 19, 2011
10. Aug 19, 2011

### disregardthat

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

11. Aug 19, 2011

### Dr. Seafood

That's what I said.

12. Aug 19, 2011

### disregardthat

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