Question

prove that
a^2+b^2/2 greater than or equal to (a+b/2)^2

Answer 1

no u cannot do so

Answer 2

sorry i didn't read carefully

Answer 3

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

Answer 4

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

Answer 5

true

Answer 6

can u solve it...

Answer 7

Yes.

Answer 8

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:\[(a^2+b^2)/2\ge (a^2+2ab+b^2)/4\]

Answer 9

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

Answer 10

is it an exclamation mark after b+?

Answer 11

sorry it is 1

Answer 12

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

Answer 13

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

Answer 14

explain the second step

Answer 15

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

Answer 16

thanks for helping anwar