Question

Dacă 0 < a <b sa se arate ca : a) ma-mh< sau egal cu (b-a)²/4a. b) ma-mg< sau egal cu (b-a)²/8a repede, e urgent ,va roog​

Answer 1

Explicație pas cu pas:

0 < a < b

$m_{a} = \frac{a + b}{2} \\ m_{g} = \sqrt{ab} \ \ \ \\ m_{h} = \frac{2ab}{a + b}$

a)

$m_{a} - m_{h} = \frac{a + b}{2} - \frac{2ab}{a + b} = \frac{ {(a + b)}^{2} - 4ab}{2(a + b)} = \\ = \frac{ {a}^{2} + 2ab + {b}^{2} - 4ab}{2(a + b)} = \frac{{b}^{2} - 2ab + {a}^{2}}{2(a + b)} = \frac{{(b - a)}^{2}}{2a + 2b}$

$b > a \implies \frac{{(b - a)}^{2}}{2a + 2b}\leqslant \frac{ {(b - a)}^{2} }{2a + 2a} = \frac{ {(b - a)}^{2} }{4a} \\ \iff \boxed {m_{a} - m_{h} \leqslant \frac{ {(b - a)}^{2} }{4a}}$

b)

$a < b \iff \sqrt{a} < \sqrt{b} \\ 2\sqrt{a} \leqslant \sqrt{a} + \sqrt{b} \iff \frac{^{ \sqrt{b} - \sqrt{a} )} 1}{\sqrt{a} + \sqrt{b}} \leqslant \frac{1}{2 \sqrt{a} } \\ \frac{\sqrt{b} - \sqrt{a} }{b - a} \leqslant \frac{1}{\sqrt{4a} } \iff {\Big(\frac{\sqrt{b} - \sqrt{a} }{b - a}\Big)}^{2} \leqslant {\Big(\frac{1}{\sqrt{4a} }\Big)}^{2} \\ \frac{a + b - 2 \sqrt{ab} }{ {(b - a)}^{2} } \leqslant \frac{1}{4a} \iff a + b - 2 \sqrt{ab} \leqslant \frac{{(b - a)}^{2}}{4a} \\ \frac{a + b}{2} - \sqrt{ab} \leqslant \frac{ {(b - a)}^{2} }{8a} \iff \boxed{ m_{a} - m_{g} \leqslant \frac{ {(b - a)}^{2} }{8a}}$