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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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no u cannot do so
sorry i didn't read carefully
hint given is (a+!)^2+(b+1)^2 is greater than equal to 0

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Other answers:

Is it \(\large \frac{a^2+b^2}{2}\ge (\frac{a+b}{2})^2\)?
true
can u solve it...
Yes.
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:\[(a^2+b^2)/2\ge (a^2+2ab+b^2)/4\]
14yamaka in my book he has used (a+1)^2+(b+!)^2>/=0
is it an exclamation mark after b+?
sorry it is 1
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 \((\frac{a^2}{2}+\frac{b^2}{2}-ab)\ge 0.\) \[(\frac{a^2}{2}+\frac{b^2}{2}-ab)=\frac{1}{2}(a-b)^2\ge 0.\]
Ask if you didn't get any of the steps.
explain the second step
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 \[a^2+b^2\ge a^2+b^2-(\frac{a^2}{2}+\frac{b^2}{2}-ab) \implies -(\frac{a^2}{2}+\frac{b^2}{2}-ab)\le 0 \implies \cdots \]
thanks for helping anwar

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