A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
prove that
a^2+b^2/2 greater than or equal to (a+b/2)^2
anonymous
 4 years ago
prove that a^2+b^2/2 greater than or equal to (a+b/2)^2

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry i didn't read carefully

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hint given is (a+!)^2+(b+1)^2 is greater than equal to 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is it \(\large \frac{a^2+b^2}{2}\ge (\frac{a+b}{2})^2\)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0uh....im probably way off but.... if everything is broken down, 2ab is less than or equal to zero and because the equation turns out like this:\[(a^2+b^2)/2\ge (a^2+2ab+b^2)/4\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.014yamaka in my book he has used (a+1)^2+(b+!)^2>/=0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is it an exclamation mark after b+?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Note that \(\large (\frac{a+b}{2})^2=\frac{a^2}{4}+\frac{b^2}{4}+\frac{ab}{2}\). The inequality can then be written as: \[a^2+b^2\ge \frac{a^2}{2}+\frac{b^2}{2}+ab=a^2+b^2(\frac{a^2}{2}+\frac{b^2}{2}ab).\] From here it would enough to show that \((\frac{a^2}{2}+\frac{b^2}{2}ab)\le 0\) or \((\frac{a^2}{2}+\frac{b^2}{2}ab)\ge 0.\) \[(\frac{a^2}{2}+\frac{b^2}{2}ab)=\frac{1}{2}(ab)^2\ge 0.\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ask if you didn't get any of the steps.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0explain the second step

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I multiplied both sides by 2 first then I wrote \[\frac{a^2}{2}+\frac{b^2}{2}+ab=a^2\frac{a^2}{2}+b^2\frac{b^2}{2}+ab=a^2+b^2(\frac{a^2}{2}+\frac{b^2}{2}ab).\] You can then write the inequality as \[a^2+b^2\ge a^2+b^2(\frac{a^2}{2}+\frac{b^2}{2}ab) \implies (\frac{a^2}{2}+\frac{b^2}{2}ab)\le 0 \implies \cdots \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks for helping anwar
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.