The question I am attempting is broken up into 2 parts:
a) Prove that $(1+2+...+n)^2 = 1^3 + 2^3 +...+n^3$
b)Find a proof for the Bernoulli inequality:
$$(1+x)^n \geq1+nx \space \text{for all}\space x \in R ,x\geq -1\space \text{and}\space n \in N,n \geq 2$$
Proof:
Proof will be by the principle of mathematical induction.
Base case $n=0$ :
$$0^2=0^3$$
Assume equation is true when $n=k$
Then $(1+2+...k+k+1)^2 = 1(1+2+...k+1) + 2(1+2+...k+1)+...k(1+2+...k+1)+(k+1)(1+2+...k+1)$
Factoring, we get$$(1+2+...k+1)^2=(\sum\limits_{r=1}^k r )(\sum\limits_{r=1}^{k+1} r) + (k
+1)(\sum\limits_{r=1}^{k+1} r)$$
Expanding $$(\sum\limits_{r=1}^k r )(\sum\limits_{r=1}^{k+1} r) + (k
+1)(\sum\limits_{r=1}^{k+1} r) =(\sum\limits_{r=1}^k r )(\sum\limits_{r=1}^{k} r) + (k
+1)(\sum\limits_{r=1}^{k} r) +(k
+1)(\sum\limits_{r=1}^{k+1} r)$$
because $\sum\limits_{r=1}^{k} r = k(k+1)/2$
then$$(\sum\limits_{r=1}^k r )(\sum\limits_{r=1}^{k} r) + (k
+1)(\sum\limits_{r=1}^{k} r) +(k
+1)(\sum\limits_{r=1}^{k+1} r) = 1^3 +2^3 +...k^3+(k+1)(\frac{(k+1)(k+2)+(k)(k+1)}{2})\space\space\space\space\space\space
= 1^3+2^3+...k^3+(k+1)^3$$Q.E.D
For b)
For base case $n=2$ $$(1+x)^2 = 1+2x+x^2\geq 1+2x $$
Suppose n=k is true, then $$(1+x)^k\geq1+kx$$
Multiplying by $(1+x)$ on both sides $$(1+x)^{k+1} \geq (1+x)(1+kx) = 1+(k+1)x+kx^2\geq 1+(k+1)x$$
As the statements holds for base case $n=2$ and $n=k$ and $n=k+1$, then the inequality is true for $n\geq 2$
Q.E.D
Are the proofs correct? Anything I could add for clarity?