Help solving an inequalities problem

[wakajo]

Hey, so I've been attempting to start on this problem for hours and I'm getting nowhere at all. Any hints or advice on how to tackle it would be greatly appreciated.

In electric circuit theory, the combined resistance R of two resistors R1 > 0 and
R2 > 0 connected in parallel obeys

(1/R)=(1/R1)+(1/R2)


Show that R < Sqrt(R1R2/2)

This proves that R cannot be large if the individual
resistances R1 and R2 are small.

 

[stapel]

...the combined resistance R of two resistors R1 > 0 and R2 > 0 connected in parallel obeys

(1/R)=(1/R1)+(1/R2)

Show that R < Sqrt(R1R2/2)

This proves that R cannot be large if the individual resistances R1 and R2 are small.

What happens when you convert the right-hand side to common denominators, combine the two terms, and then "flip" the fractions on either side of the "equals" sign?

 

[wakajo]

What happens when you convert the right-hand side to common denominators, combine the two terms, and then "flip" the fractions on either side of the "equals" sign?

so i understand that, my only question is how to begin my proof, do i start with r1r2/(r1+r2)= R and try to make that equal to the inequality? or do i have to do something with the fact that r1>0 and r2>0. Im tried setting up r1r2/(r1+r2)< sqrt(r1r2/2) but i feel like im just going in circles and not actually proving anything. Thanks so much for your time

 

[stapel]

so i understand that, my only question is how to begin my proof, do i start with r1r2/(r1+r2)= R and try to make that equal to the inequality? or do i have to do something with the fact that r1>0 and r2>0. Im tried setting up r1r2/(r1+r2)< sqrt(r1r2/2) but i feel like im just going in circles and not actually proving anything. Thanks so much for your time

Are you saying that you tried to start your proof with the statement that you are needing to prove? If so, then this would be invalid. You need to start with what you've been given, and then work logically to what you're trying to prove. The statement to be proved should come at the end of the proof, not at the beginning.

Meanwhile, it sounds from your reply that you have done a fair amount of work on this, including the "start" at which I'd hinted. Please reply showing your work and reasoning so far. Thank you!

 

[wakajo]

Ok yea that's what I thought. I tried setting the equation up with r1r1/(r1+r2)>0 and tried playing with it to make it look like the inequality I need to prove. However I spent hours getting nowhere. I don't know what work to show you as it all is just manipulations of an inequality hoping that somehow it would work out. what I really don't understand is how to go from from the original equation to an inequality

 

[Ishuda]

Ok yea that's what I thought. I tried setting the equation up with r1r1/(r1+r2)>0 and tried playing with it to make it look like the inequality I need to prove. However I spent hours getting nowhere. I don't know what work to show you as it all is just manipulations of an inequality hoping that somehow it would work out. what I really don't understand is how to go from from the original equation to an inequality

R = \(\displaystyle \frac{R1\space R2}{R1\space +\space R2}\)
or
R² = \(\displaystyle \frac{R1^2 R2^2}{(R1\space +\space R2)^2}\) = \(\displaystyle \frac{R1^2 R2^2}{R1^2+\space 2\space R1\space R2 +\space R2^2}\)

Can you go from there? if you are still stuck, highlight the area between the lines of *'s for a hint:
*******************************
Hint: If a and c are non-negative and b is positive \(\displaystyle \frac{a}{b+c}\le \frac{a}{b}\).
*******************************

 

[wakajo]

R = \(\displaystyle \frac{R1\space R2}{R1\space +\space R2}\)
or
R² = \(\displaystyle \frac{R1^2 R2^2}{(R1\space +\space R2)^2}\) = \(\displaystyle \frac{R1^2 R2^2}{R1^2+\space 2\space R1\space R2 +\space R2^2}\)

Can you go from there? if you are still stuck, highlight the area between the lines of *'s for a hint:
*******************************
Hint: If a and c are non-negative and b is positive \(\displaystyle \frac{a}{b+c}\le \frac{a}{b}\).
*******************************

Thank you so much, I just never thought to square it like that!