## mathmath333 one year ago Question

1. mathmath333

\large \color{black}{\begin{align} & \{a,b\}\in \mathbb{Z} ,\ \ \ a,b >2,\ a>b\ \hspace{.33em}\\~\\ & \normalsize \text{which of the following will be always true ?} \hspace{.33em}\\~\\ & a.)\ \dfrac{a+b}{2}>\dfrac{a-b}{2} \hspace{.33em}\\~\\ & b.)\ \dfrac{a+b}{2}\geq \dfrac{a-b}{2} \hspace{.33em}\\~\\ & c.)\ a^2-b^2<a^3-b^3\hspace{.33em}\\~\\ & d.)\ \normalsize \text{Both a and c} \hspace{.33em}\\~\\ & e.)\ \normalsize \text{ a, b and c} \hspace{.33em}\\~\\ \end{align}}

2. anonymous

B is confusing, but I thinks it's true The answer is E since we are dealing with positive numbers and bigger than one

3. mathmath333

but in the book it is given as option d.)

4. anonymous

Not too good at calc but here it goes... What I did was, was I basically picked random numbers that are greater than 2. (I did multiple trials). Trail 1: a = 8 while b = 6 A. True B. True C. True Trial 2: a = 12 while b = 4 A. True B. True C. True So I would have to agree with @Ahmad-nedal that the answer is E. However, I am not in calc therefore I don't serve much of a purpose. Perhaps it is d because of the {a,b} E Z.

5. anonymous

for a and b note that $$a+b$$ is strictly greater than $$a-b$$

6. anonymous

I understand the reason, some books deal with ineaqalities from different aspects. Although the equality stated in part B will not be achieved, but the argument still holds for any a,b in z that satsifies the constraints given. So I insist on my opinion

7. anonymous

for c you can divide both sides of inequality by $$a-b$$ which is a positive number and get$a+b<a^2+ab+b^2$which is true clearly

8. mathmath333

is the answer option e.) ?

9. anonymous

Yes

10. anonymous

it's d

11. mathmath333

ok thnx

12. anonymous

It is given that a and b are integers greater than 2 and that a is always greater than b. You can choose a number. Let as say a=4 and b=3 For a)= $\frac{ 4+3 }{ 2 }>\frac{ 4-3 }{ 2 }=\frac{ 7 }{ 2 }>\frac{ 1 }{ 2 }$ Similarly at c) $4^{2}-3^{2}<4^{3}-3^{3}=7<37$ Therefore both a and c are true.