Question

Se considera numerele reale a=√6-√3 si b=√6+√3. Aratati ca:
$\frac{4}{3}< \frac{1}{a}+ \frac{1}{b}< \frac{5}{3}.$

Answer 1

$\frac{4}{3}$<$\frac{1}{ \sqrt{6}- \sqrt{3}}+\frac{1}{ \sqrt{6}+ \sqrt{3}}$ ⇒ $\frac{4}{3}< \frac{ \sqrt{6}+\sqrt{3}+\sqrt{6}-\sqrt{3}}{6-3}$ ⇒ $\frac{4}{3}< \frac{2 \sqrt{6} }{3}$ ⇒ 2<√6 ridicam la patrat ⇒ 4<6 adevarat
Pentru a arata ca este mai mic decat 5/3 am calculat mai sus ca 1/a + 1/b = 2√6/3 <5/3 ⇒ 2√6<5 ridicam la patrat ⇒ 24<25 adevarat