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

[mathlover1]

For all positive real numbers $$x,y$$ prove that:

$$\frac{1}{1+\sqrt{x}}+\frac{1}{1+\sqrt{y}} \geq \frac{2\sqrt{2}}{1+\sqrt{2}}$$

 

[ohubrismine]

For all positive real numbers $$x,y$$ prove that:

$$\frac{1}{1+\sqrt{x}}+\frac{1}{1+\sqrt{y}} \geq \frac{2\sqrt{2}}{1+\sqrt{2}}$$

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.

 

[mathlover1]

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

 

[Tedjn]

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

 

[mathlover1]

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!

 

[Mark44]

Find the minimum value of
$$f(x) = \frac{1}{1 + \sqrt{x} } + \frac{1}{1 + \sqrt{1 - x}}$$

This involves finding the critical points.