Proving the Inequality |x+y|^p \leq 2^p(|x|^p+|y|^p)

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In summary, the conversation was about proving the inequality |x+y|^p \leq 2^p(|x|^p+|y|^p) for any positive integer p and real numbers x,y. The solution involved using the triangle inequality and induction to show that the inequality holds true for p+1 if it is already true for p.
  • #1
quasar987
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[SOLVED] Help me prove this inequality

Homework Statement


The inequality in question is

[tex]|x+y|^p \leq 2^p(|x|^p+|y|^p)[/tex]

for any positive integer p and real numbers x,y.

The Attempt at a Solution


For p=1, it is weaker than the triangle inequality.

Suppose it is true for p, and let's try to show this implies it's true for p+1.

[tex]|x+y|^{p+1}=|x+y||x+y|^p\leq |x+y|2^p(|x|^p+|y|^p)[/tex]

And basically, here I've tried using the triangle inequality on |x+y| but the most "reduced form" I got is I arrived at the conclusion that the inquality was true iff

[tex]|x||y|(|x|^p+|y|^p)\leq |x|^{p+1}+|y|^{p+1}[/tex]
 
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  • #2
Let [tex]a,b\geq 0[/tex] we want to show [tex](a+b)^n \leq 2^n(a^n+b^n)[/tex]. First accept induction. Then [tex](a+b)^{n+1} = (a+b)^n(a+b)\leq 2^n(a^n+b^n)(a+b) = 2^n(a^{n+1}+b^{n+1}+ab^n+a^nb)[/tex]. But [tex]2^n(a^{n+1}+b^{n+1}+ab^n+a^nb)\leq 2^{n+1}(a^{n+1}+b^{n+1})[/tex] iff [tex]a^{n+1}+b^{n+1}\leq 2(a^{n+1}+b^{n+1}+a^nb+ab^n)[/tex] iff [tex]a^{n+1}+b^{n+1}\geq ab^n+a^nb[/tex] iff [tex](a-b)(a^n-b^n)\geq 0[/tex] but that is true because [tex](a-b)(a^n-b^n) = (a-b)^2(a^{n-1}b+...+ab^{n-1})\geq 0[/tex].
 
  • #3
Kummer said:
But [tex]2^n(a^{n+1}+b^{n+1}+ab^n+a^nb)\leq 2^{n+1}(a^{n+1}+b^{n+1})[/tex] iff [tex]a^{n+1}+b^{n+1}\leq 2(a^{n+1}+b^{n+1}+a^nb+ab^n)[/tex]

Ok, this is just a typo probably because you fall back on your feet a few lines later with

[tex]a^{n+1}+b^{n+1}\geq ab^n+a^nb[/tex]

Good work, thanks Kummer.
 
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  • #4
How do I add [SOLVED] to the title?
 
  • #5
quasar987 said:
How do I add [SOLVED] to the title?

Click on "thread tools".
 

1. What is an inequality?

An inequality is a mathematical statement that compares two values or expressions using symbols such as <, >, ≤, or ≥. It indicates that one value is either smaller or larger than the other.

2. How do I prove an inequality?

To prove an inequality, you need to demonstrate that the statement is true for all possible values of the variables involved. This can be done using various mathematical techniques such as algebra, calculus, or logic.

3. Why is proving an inequality important?

Proving an inequality is important because it allows us to establish the truth or validity of a mathematical statement. Inequalities are used in many fields of science, such as economics and physics, to describe relationships between variables.

4. What are some common techniques used to prove inequalities?

Some common techniques used to prove inequalities include substitution, factoring, using the properties of inequalities, and using mathematical induction. The choice of technique depends on the specific inequality and the variables involved.

5. Can you provide an example of proving an inequality?

Sure, for example, to prove that 2x + 3 < 7 for all values of x, we can use the property of inequalities that states: if a < b and c < d, then a + c < b + d. Therefore, if x < 2, then 2x < 2(2) = 4. Similarly, if x < 1, then 3 < 3x. Combining these two inequalities, we get 2x + 3 < 4 + 3x. Since x < 1 and x < 2, we can say that 3x < 3 and 4 < 7. Using the property of inequalities, we can conclude that 2x + 3 < 7 for all values of x.

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