Here's the problem: Sq root(ab) ≤ (a+b)/2 So I'm stuck here, how do I prove that a and b are positive real numbers? Should I go with a counterexample?
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a= -1 b= -1 sqrt(ab)=sqrt(1)=1 (a+b)/2 = (-1-1)/2 = -1 sqrt(ab)>(a+b)/2
Originally Posted by bxuan Here's the problem: Sq root(ab) ≤ (a+b)/2 prove that a and b are positive real numbers? It is true $\displaystyle (p-q)^2\ge 0$ so $\displaystyle p^2+q^2\ge 2pq$. If $\displaystyle a>~\&~b>0$ then let $\displaystyle p=\sqrt{a}~\&~q=\sqrt{b}$.
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