Solving xy ≤ (x^p/p) + (y^q/q) Problem

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In summary, the conversation discusses a problem that involves showing an inequality for positive numbers x and y, given a condition on the exponents p and q. The logarithm function is used to simplify the problem, and the potential use of derivatives is suggested. The final solution involves using the Arithmetic Mean-Geometric Mean inequality.
  • #1
ELESSAR TELKONT
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My problem is the next:

Show that [tex]\forall x,y\geq 0[/tex] and [tex]\frac{1}{p}+\frac{1}{q}=1[/tex] we have that [tex]xy\leq\frac{x^{p}}{p}+\frac{y^{q}}{q}[/tex].

I use the logarithm function because it is concave and grows monotonically and then

[tex]\ln{xy}=\frac{\ln{x^{p}}}{p}+\frac{\ln{y^{q}}}{q}[/tex]

the problem is I don't know to justify the next step (or is justified yet?)

[tex]\ln{xy}=\frac{\ln{x^{p}}}{p}+\frac{\ln{y^{q}}}{q}\leq \ln\left(\frac{x^{p}}{p}+\frac{y^{q}}{q}\right)[/tex].

That's all, I wait for your help.
 
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  • #2
Derivatives are often good at proving inequalities. e.g. you can prove
f(x) < g(x)
for all positive x as follows:
. f(0) < g(0)
. f'(x) < g'(x) for all positive x

(I don't know if they will be useful here, but it's something I'd try)
 
  • #3
Hi
This may be brute force but it looks like it works:
Use Arithmetic Mean>= Geometric Mean.
1/p+1/q=1 implies p+q=p.q
Also x^p/p+y^q/q=(q.x^p+p.y^q)/(p.q)
Now AM>=GM implies
(q instances of x^p plus p instances of y^q)/(p+q)>={x^(p.q).y^(q.p)}^(1/(p+q))
LHS is just (p.q).{x^p/p+y^q/q}/(p+q)=x^p/p+y^q/q
RHS is x.y

Hope this helps
 
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  • #4
Hurkyl, Derivatives only help easily in inequalities of 1 variable :( Even if he expressed y as a function of x, the condition on p and q pose difficulty.
 

1. What is the purpose of solving the inequality xy ≤ (x^p/p) + (y^q/q)?

The purpose of solving this inequality is to find the set of values for x and y that satisfy the given equation. This can help determine the range of possible solutions and provide insight into the relationship between the variables.

2. How do you approach solving this type of inequality problem?

One approach is to graph the equation and identify the points where the graph intersects the x and y axes. These points can then be used to create a table of values to determine the range of solutions. Another approach is to manipulate the equation algebraically to isolate the variables and determine the range of values that satisfy the inequality.

3. Are there any specific values for p and q that make solving this inequality easier?

Yes, if p and q are integers, it can be easier to solve this inequality as it allows for easier manipulation of the equation. Additionally, if p and q are both positive, the inequality will be easier to solve as the exponents will not change the direction of the inequality.

4. What are some common mistakes to avoid when solving this type of problem?

Some common mistakes include forgetting to distribute the exponents when simplifying the equation, not properly isolating the variables on one side of the inequality, and incorrectly graphing the equation. It is also important to pay attention to the signs of the coefficients and exponents when manipulating the equation.

5. How can solving this inequality be applied in real-world situations?

Solving this inequality can be useful in various fields such as economics, physics, and engineering. It can help determine the optimal values for variables in a given situation, such as maximizing profit or minimizing cost in a business setting, or finding the maximum efficiency in a physical system. It can also be used to model and predict various phenomena, such as population growth or the spread of diseases.

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