Question

va rog urgent!!!!!!!!!!!

Answer 1

Media armonica:

$m_h=\frac{2ab}{a+b}$

Media aritmetica:

$m_a=\frac{a+b}{2}$

Media geometrica:

$m_g=\sqrt{ab}$

a)

$\frac{a+b}{2} -\frac{2ab}{a+b} \leq \frac{(b-a)^2}{4a} \\\\\frac{(a+b)^2-4ab}{2(a+b)} \leq \frac{(b-a)^2}{4a} \\\\\frac{a^2-2ab+b^2}{2(a+b)} \leq \frac{(b-a)^2}{4a} \\\\\frac{(b-a)^2}{2(a+b)} \leq \frac{(b-a)^2}{4a}$

Comparam 2(a+b) cu 4a

2a+2b   4a

Stim ca b>a

2a+2b    4a    |-2a

2b    2a

b>a⇒   2b>2a⇒$\frac{(b-a)^2}{2(a+b)} \leq \frac{(b-a)^2}{4a}$

b)

$\frac{a+b}{2} -\sqrt{ab} \leq \frac{(b-a)^2}{8a} \\\\\frac{a+b-2\sqrt{ab} }{2} \leq \frac{(b-a)^2}{8a} \\\\\frac{(\sqrt{b}-\sqrt{a} )^2 }{2} \leq \frac{(b-a)^2}{8a} \ \ \ | \cdot 2$

$(\sqrt{b} -\sqrt{a} )^2\leq \frac{(b-a)^2}{4a} \\\\\frac{(\sqrt{b} -\sqrt{a} )^2}{(b-a)^2} \leq \frac{1}{4a}$

$\frac{\sqrt{b} -\sqrt{a} }{b-a} \leq \frac{1}{2\sqrt{a} }$

$\frac{b-a}{(b-a)(\sqrt{b}+\sqrt{a} ) } \leq \frac{1}{2\sqrt{a} } \\\\\frac{1}{\sqrt{b}+\sqrt{a} } \leq \frac{1}{2\sqrt{a} }\\\\\ comparam \ \sqrt{b} +\sqrt{a}\ si \ 2\sqrt{a}\ \ \ \ | - \sqrt{a} \\\\ \sqrt{b} \geq \sqrt{a } \ deoarece \ b > a\ rezulta\ ca\ \frac{1}{\sqrt{b}+\sqrt{a} } \leq \frac{1}{2\sqrt{a} }$

Un alt exercitiu gasesti aici: https://brainly.ro/tema/2809690

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