Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
@mukushla
Prove \(ab + bc + ac \ge \sqrt{3abc \left(a+b+c\right)}\).
 one year ago
 one year ago
@mukushla Prove \(ab + bc + ac \ge \sqrt{3abc \left(a+b+c\right)}\).
 one year ago
 one year ago

This Question is Closed

mukushlaBest ResponseYou've already chosen the best response.1
could u plz typr it again?
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
i am sorry, i redid it. refresh.
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
it just needed squaring and then it becomes, \[a^2b^2 + c^2b^2 + a^2c^2 \ge abc(a+b+c)\] and then a simple complete square. i act so foolishly sometimes, sorry for troubling you mukushla
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
squaring both sides \[a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)≥ 3abc(a+b+c)\\ a^2b^2+b^2c^2+c^2a^2abc(a+b+c)\ge 0 \\2a^2b^2+2b^2c^2+2c^2a^22abc(a+b+c)\ge 0\\a^2b^22a^2bc+a^2c^2+a^2b^22ab^2c+b^2c^2+a^2c^22abc^2+b^2c^2\ge0\\ (ab−bc)^2+(bc−ca)^2+(ca−ab)^2\ge0\]
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.