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Generally we have
\[  \frac{a^{2}+b^{2}}{2}\geq ab.\]
Equality holds onl if $a=b$. The proof is as follows:
\[
  \frac{a^{2}+b^{2}}{2}= ab \mbox{ thus } 0=a^{2}-2ab-b^{2} = (a-b)^{2}
\]



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