## anonymous 4 years ago prove that a^2+b^2/2 greater than or equal to (a+b/2)^2

1. anonymous

no u cannot do so

2. anonymous

3. anonymous

hint given is (a+!)^2+(b+1)^2 is greater than equal to 0

4. anonymous

Is it $$\large \frac{a^2+b^2}{2}\ge (\frac{a+b}{2})^2$$?

5. anonymous

true

6. anonymous

can u solve it...

7. anonymous

Yes.

8. anonymous

uh....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:$(a^2+b^2)/2\ge (a^2+2ab+b^2)/4$

9. anonymous

14yamaka in my book he has used (a+1)^2+(b+!)^2>/=0

10. anonymous

is it an exclamation mark after b+?

11. anonymous

sorry it is 1

12. anonymous

Note 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 $$(\frac{a^2}{2}+\frac{b^2}{2}-ab)\ge 0.$$ $(\frac{a^2}{2}+\frac{b^2}{2}-ab)=\frac{1}{2}(a-b)^2\ge 0.$

13. anonymous

Ask if you didn't get any of the steps.

14. anonymous

explain the second step

15. anonymous

I 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 $a^2+b^2\ge a^2+b^2-(\frac{a^2}{2}+\frac{b^2}{2}-ab) \implies -(\frac{a^2}{2}+\frac{b^2}{2}-ab)\le 0 \implies \cdots$

16. anonymous

thanks for helping anwar