Question

a+b=1
Dem: a^3+b^3>=1/4

Answer 1

    
[tex]\displaystyle a+b=1 \\ a^3+b^3= \frac{1}{4} \\ \text{--------------------} \\ b=1-a \\ a^3+(1-a)^3 =\frac{1}{4} \\ a^3 + 1 -3a + 3a^2 -a^3=\frac{1}{4} \\ 3a^2-3a+1=\frac{1}{4} ~~~~~| \times 4 \\ 12a^2 - 12a + 4=1 \\ 12a^2 - 12a + 3=0 ~~~~~|: 3 \\ 4a^2 - 4a +1=0 [/tex]

[tex]\displaystyle 4a^2 - 4a +1=0 \\ \\ a_{_{12}}= \frac{4\pm \sqrt{16-4\times 4} }{8}= \frac{4\pm \sqrt{16-16} }{8} = \frac{4}{8} = \frac{1}{2} \\ a =\boxed{ \frac{1}{2} }~~~~~radacina~~dubla \\ \\ b= 1-a = 1-\frac{1}{2} = \boxed{ \frac{1}{2} }~~~~~radacina~~dubla \\ \\ \text{Solutia problemei:} \\ \\ \boxed{\boxed{a = b = \frac{1}{2} }}[/tex]