A community for students.
Here's the question you clicked on:
 0 viewing

This Question is Closed

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.1could u plz typr it again?

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.1i am sorry, i redid it. refresh.

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.1it 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

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.1squaring both sides \[a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)≥ 3abc(a+b+c)\\ a^2b^2+b^2c^2+c^2a^2abc(a+b+c)\ge 0 \\2a^2b^2+2b^2c^2+2c^2a^22abc(a+b+c)\ge 0\\a^2b^22a^2bc+a^2c^2+a^2b^22ab^2c+b^2c^2+a^2c^22abc^2+b^2c^2\ge0\\ (ab−bc)^2+(bc−ca)^2+(ca−ab)^2\ge0\]
Ask your own question
Sign UpFind more explanations on OpenStudy
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.