Question

Daca a,b>0 si a²+b²=9,demonstrati ca:
$\frac{2 \sqrt{2} }{a} + \frac{1}{b} \geq \sqrt{3}$

Answer 1

$\displaystyle Din~inegalitatea~lui~Holder~avem: \\ \\ \left(\frac{2 \sqrt{2}}{a} + \frac{1}{b} \right) \left(\frac{2 \sqrt{2}}{a} + \frac{1}{b} \right)(a^2+b^2) \geq \\ \\ \geq \left( \sqrt[3]{\frac{2 \sqrt{2}}{a} \cdot \frac{2 \sqrt{2}}{a} \cdot a^2}+ \sqrt[3]{\frac{1}{b} \cdot \frac{1}{b} \cdot b^2} \right)^3= \\ \\ =( \sqrt[3]{8}+ \sqrt[3]{1})^3= \\ \\ =(2+1)^3= \\ \\ =27.$

$\displaystyle Deci~\left(\frac{2 \sqrt{2}}{a} + \frac{1}{b} \right)^2(a^2+b^2) \geq 27 \Leftrightarrow \left(\frac{2 \sqrt{2}}{a} + \frac{1}{b} \right)^2 \cdot 9 \geq 27. \\ \\ Deci ~ \left(\frac{2 \sqrt{2}}{a} + \frac{1}{b} \right)^2 \geq 3,~adica~\left(\frac{2 \sqrt{2}}{a} + \frac{1}{b} \right) \geq \sqrt{3}.$