## Online Course Discussion Forum

### MC III Algebra Ch. 5 Help

Re: MC III Algebra Ch. 5 Help

I have been traveling, so responded late.

Here is a more complete solution for 27:

The following inequalities are true for all real numbers $a,b,c,d$: $\def\arraystretch{2.2} \begin{array}{rcl} 2a^2 + \dfrac{b^2}{2} \geq 2ab, \\ \dfrac{3}{2}b^2 + \dfrac{3}{2} c^2 \geq 3bc, \\ \dfrac{c^2}{2} + 2d^2 \geq 2cd. \end{array}$ Adding the inequalities, $2a^2 + 2b^2 + 2c^2 + 2d^2 \geq 2ab + 3bc + 2cd,$ and equality occurs when $b=c=2a=2d$. Therefore the minimum value $M=2$.

The coefficients were found by using the parameter $k$ like suggested before.

For 28, what you did is pretty good; to estimate $a_{91}$, the following can be done:

It is clear that $\{a_k\}$ is an increasing sequence, and for all $k\geq 1$, $a_{k+1} - a_{k} = a_k^2 \geq a_1^2 = \frac{1}{100},$ hence $a_{91} = (a_{91} - a_{90}) +(a_{90} - a_{89}) + \cdots + (a_2 - a_1) + a_1 > 90\cdot \frac{1}{100} + \frac{1}{10} = 1.$