Question

Prove a desigualdade
$(a + b)( \frac{1}{a} + \frac{4}{b}) \geqslant 9$
quando a>0 e b>0.Determine quando ocorre a igualdade​

Answer 1

Deseja-se provar que:

$\mathsf{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{4}{b}\right) \geq 9}$

com $\mathsf{a>0}$ e $\mathsf{b>0.}$

Ademais, pede-se para determinar quando a igualdade ocorre.

Prova:

Vamos partir do lado esquerdo da desigualdade e chegar ao segundo.  Para isso, iremos utilizar a desigualdade de Cauchy- Schwarz.

$\dotfill$

Desigualdade de Cauchy- Schwarz: Se a, b, c e d são números reais, então:

$\boxed{\mathsf{\left(a^2+b^2\right)(c^2+d^2)\geq{\left(ac+bd\right)}^2}}$

$\dotfill$

Daí:

$\mathsf{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{4}{b}\right)}\geq\\\\ \geq\mathsf{{\left(\sqrt{a}\cdot\sqrt{\dfrac{1}{a}}+\sqrt{b}\cdot\sqrt{\dfrac{4}{b}}\right)}^2}\\\\ \geq \mathsf{{\left(\sqrt{a\cdot\dfrac{1}{a}}+\sqrt{b\cdot\dfrac{4}{b}}\right)}^2}\\\\\geq\mathsf{{\left(\sqrt{1}+\sqrt{4}\right)}^2}\\\\\geq\mathsf{{(1+2)}^2}\\\\\geq \mathsf{3^2}\\\\\geq \mathsf{9}\qquad\qquad\blacksquare$

Quando ocorre a igualdade?

$\mathsf{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{4}{b}\right) = 9} \iff\\\\\iff \mathsf{(a+b)(b+4a)=9ab}\\\\\iff\mathsf{ab+4a^2+b^2+4ab-9ab=0}\\\\\iff\mathsf{4a^2+b^2-4ab=0}\\\\\iff\mathsf{(2a)^2+b^2-2\cdot(2a)\cdot (b)=0}\\\\\iff\mathsf{(2a-b)^2=0} \\\\\iff\mathsf{2a=b}$

Logo, a igualdade ocorre quando $\mathsf{b=2a.}$