What is the Minimum Value of f(x) for Positive Real Numbers x and y?

Click For Summary

Discussion Overview

The discussion revolves around proving an inequality involving positive real numbers x and y, specifically the expression \(\frac{1}{1+\sqrt{x}}+\frac{1}{1+\sqrt{y}} \geq \frac{2\sqrt{2}}{1+\sqrt{2}}\). Participants explore the conditions under which this inequality holds, particularly focusing on the case where \(x + y = 1\).

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that the inequality cannot be proven, citing the case where \(x = y = 1\) leading to a left-hand side value of 1, which is less than the right-hand side.
  • Others clarify that the values of \(x\) and \(y\) are constrained by the condition \(x + y = 1\), suggesting that this may affect the validity of the inequality.
  • A participant proposes to find the minimum value of the function \(f(x) = \frac{1}{1 + \sqrt{x}} + \frac{1}{1 + \sqrt{1 - x}}\) as a way to approach the problem.
  • There is a call for collaboration, with one participant expressing a desire for assistance in proving the inequality.

Areas of Agreement / Disagreement

Participants do not reach a consensus; some believe the inequality is unprovable under certain conditions, while others suggest that the constraints on \(x\) and \(y\) may allow for a proof.

Contextual Notes

The discussion is limited by the assumption that \(x + y = 1\), which may influence the behavior of the function being analyzed. There are also unresolved mathematical steps regarding the critical points of the function.

mathlover1
Messages
8
Reaction score
0
For all positive real numbers [tex]x,y[/tex] prove that:

[tex]\frac{1}{1+\sqrt{x}}+\frac{1}{1+\sqrt{y}} \geq \frac{2\sqrt{2}}{1+\sqrt{2}}[/tex]
 
Mathematics news on Phys.org


mathlover1 said:
For all positive real numbers [tex]x,y[/tex] prove that:

[tex]\frac{1}{1+\sqrt{x}}+\frac{1}{1+\sqrt{y}} \geq \frac{2\sqrt{2}}{1+\sqrt{2}}[/tex]

No proof is possible.
Let x=y=1, then left hand side is equal to 1 which is strictly less than the right hand side.
 


ohubrismine said:
No proof is possible.
Let x=y=1, then left hand side is equal to 1 which is strictly less than the right hand side.

I forgot to say that x,y are numbers such that x+y=1
 


Since x+y=1, y=1-x. Now, you are looking for the minimum. How would you do it?
 


Tedjn said:
Since x+y=1, y=1-x. Now, you are looking for the minimum. How would you do it?
That's why i am asking to you guys!
This inequality is correct .. just need to be proven!
 


Find the minimum value of
[tex]f(x) = \frac{1}{1 + \sqrt{x} } + \frac{1}{1 + \sqrt{1 - x}}[/tex]

This involves finding the critical points.
 

Similar threads

MHB Inequality of positive real numbers
  • · Replies 1 ·
Replies
1
Views
1K
MHB Inequality involving positive real numbers
  • · Replies 1 ·
Replies
1
Views
1K
MHB Find real solutions to a system
  • · Replies 6 ·
Replies
6
Views
1K
MHB Solve Real Number $x$ That Satisfies Equation
  • · Replies 4 ·
Replies
4
Views
1K
MHB Challenge Problem #7: Σ(x/(y^3+2))≥1
  • · Replies 2 ·
Replies
2
Views
2K
MHB Inequality with positive real numbers a and b
  • · Replies 1 ·
Replies
1
Views
1K
Undergrad A question about Young's inequality and complex numbers
  • · Replies 13 ·
Replies
13
Views
2K
MHB What is the Sum of x, y, and z in a Non-Negative Real Number System?
  • · Replies 4 ·
Replies
4
Views
2K
MHB Equation 2: Prove that ##x^2+2x\sqrt x+3x+2\sqrt x+1=0##
  • · Replies 4 ·
Replies
4
Views
2K
MHB Can inequality be proven with positive real numbers and fractions?
  • · Replies 4 ·
Replies
4
Views
2K