For which case would the applied force be greater?

[haruspex]

how should I do that? Can you give an example?

Assuming x>y>0, which is larger, $x-y$ or $x+y$?
What about $\frac 1{x-y}$ and $\frac 1{x+y}$?

 

[paulimerci]

Assuming x>y>0, which is larger, $x-y$ or $x+y$?
What about $\frac 1{x-y}$ and $\frac 1{x+y}$?

X+y and 1/(x+y) are greater.
Thank you all for your great help!

 

[haruspex]

X+y and 1/(x+y) are greater.

3>2, so is 1/3 >1/2 or 1/2>1/3?

 

[paulimerci]

1/2>1/3

 

[haruspex]

1/2>1/3

Right, so if $x+y>x-y$ is $\frac 1{x+y}>\frac1{x-y}$ or $\frac 1{x+y}<\frac1{x-y}$.

 

[paulimerci]

1/(x+y) < 1/(x-y)

 

[haruspex]

1/(x+y) < 1/(x-y)

Right, so is F_A or F_B the greater?

 

[paulimerci]

Right, so is F_A or F_B the greater?

F_{B is greater}

 

[haruspex]

F_{B is greater}

Right. All clear now?

 

[paulimerci]

Right. All clear now?

Yes @haruspex , Thank you!

 

[PeroK]

Right, so if $x+y>x-y$ is $\frac 1{x+y}>\frac1{x-y}$ or $\frac 1{x+y}<\frac1{x-y}$.

We also need the condition $x > y$ here.

 

[haruspex]

We also need the condition $x > y$ here.

Yes, that was specified in post #31.