Question

Prove a desigualdade
$(a + b)( \frac{1}{a) + \frac{1}{b} \geqslant 4 \: quando \: a > 0$
e determine quando ocorre a igualdade ​

Answer 1

$\mathsf{(a+b)(\dfrac{1}{a}+\dfrac{1}{b})\ge4~quando~a>0}$

$\mathsf{(a+b)(\dfrac{a+b}{ab}) } \ge4 \times (ab) \\{(a + b)}^{2} \ge4ab \\ {a}^{2} + 2ab + {b}^{2} \ge4ab$

$\mathsf{a^2+2ab+b^2-2ab\ge4ab-2ab}$

$\mathsf{a^2+b^{2}\ge2ab}$

Supondo a=b

$\mathsf{a^2+a^2\ge2a.a}\\\mathsf{2a^2\ge2a^2}$