ab < (a + b)/2

[Guest]

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]

[nathi123]

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].