anonymous
  • anonymous
prove that a^2+b^2/2 greater than or equal to (a+b/2)^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
no u cannot do so
anonymous
  • anonymous
sorry i didn't read carefully
anonymous
  • anonymous
hint given is (a+!)^2+(b+1)^2 is greater than equal to 0

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

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