# Why is the line in bold valid (Picture added below)?

Jul 22, 2017

Refer to the Explanation.

#### Explanation:

So far, it has been proved, that,

${\left(1 + p\right)}^{k + 1} \ge 1 + k p + p + k {p}^{2.} \ldots \left({\ast}^{1}\right) .$

Now, note that, $p > - 1 \Rightarrow {p}^{2} > 0. \text{ Also, } k \in {\mathbb{Z}}^{+} .$

$\therefore k {p}^{2} > 0.$

Adding, on both sides of this inequality $1 + k p + p ,$ we get,

$k {p}^{2} + 1 + k p + p > 1 + k p + p \ldots \ldots \ldots \ldots . \left({\ast}^{2}\right) .$

Combining, $\left({\ast}^{1}\right) \mathmr{and} \left({\ast}^{2}\right) ,$ we have,

${\left(1 + p\right)}^{k + 1} \ge 1 + k p + p + {p}^{2} > 1 + k p + p , i . e . ,$

${\left(1 + p\right)}^{k + 1} > 1 + \left(k + 1\right) p .$

In simple form, the line in bold means that,

if, a >= b+c, &, c >0, then, if we drop $c > 0 ,$ from $b + c ,$

we still have, $a > c .$