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zmudz
 one year ago
Prove that if \(x,y,z\) are positive real numbers then the following inequality holds
\( \frac {x+y}{x^2+y^2} + \frac {y+z}{y^2+z^2} + \frac {z+x}{z^2+x^2} \leq \frac 1x + \frac 1y + \frac 1z . \)
zmudz
 one year ago
Prove that if \(x,y,z\) are positive real numbers then the following inequality holds \( \frac {x+y}{x^2+y^2} + \frac {y+z}{y^2+z^2} + \frac {z+x}{z^2+x^2} \leq \frac 1x + \frac 1y + \frac 1z . \)

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jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3Have you attempted anything already? Just so I don't go down a dead end path if you've already tried it. :)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3k, what class is this for so I know what level of math to use?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3Right now, I am going to try to get the fractions to have the same denominators, the algebra is very messy for now... but I'm going to see where it takes me. I'll do this on scrap paper, rather than on the computer. But know that I am trying to solve it. ;)

zmudz
 one year ago
Best ResponseYou've already chosen the best response.0sounds good, I appreciate this a lot!

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3hmm... this one is nasty... I'm trying a new approach... :)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3Almost have something...

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3Ahh... so close. How have your classes been proving things in precalculus?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3Because I could make an argument using around 68 cases, but it's a bit tedious... :)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3Sorry, I'm struggling a bit. I might just type up a pdf to show the solution. It's a bit messy to look at, but I can't see a simpler method yet.

zmudz
 one year ago
Best ResponseYou've already chosen the best response.0AM/GM inequality, maxima minima, induction, mean inequality chain

zmudz
 one year ago
Best ResponseYou've already chosen the best response.0take your time, thanks. pdf and any help at all is great :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry to get off topic, how do you type out math latex in a pdf?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3I use overleaf.com it's webbased and allows you to download a pdf after typing LaTeX.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3Here's my attempt, I didn't finish the proof, because my method was brute force and very tiring...
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