ab < (a + b)/2

ab < (a + b)/2

Postby Guest » Sat Sep 16, 2017 11:47 pm

How do I prove this statement using properties of numbers: If a < 0, b < 0, then [tex]\sqrt{ab} \le -\frac{a+b}{2}[/tex]





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Re: ab < (a + b)/2

Postby nathi123 » Sun Sep 17, 2017 2:14 pm

If[tex]a>0;b>0 \Rightarrow \sqrt{ab}\le \frac{a+b}{2}\Leftrightarrow \frac{a+b-2\sqrt{ab}}{2}\Leftrightarrow \frac{(\sqrt{a}-\sqrt{b})^{2}}{2}\ge 0[/tex]-
possible for each of a>o;b>0.
Now if a<0;b<0 [tex]\Rightarrow -a>0 ; -b>0 \Rightarrow \frac{-a + (-b) }{2}\ge \sqrt{(-a)(-b)}\Leftrightarrow - \frac{a+b}{2}\ge\sqrt{ab}[/tex].

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