Proof related with mathematical induction I tried to prove this claim using mathematical induction.
$ a^2 + 15a + 5 ≤ 21 a^2 $ $\;\; ∀a∈\mathbb Z^+$
The way is as the following:
Basis: for a = 1 is true since 21 = 21
Inductive step: If $P(t)$ then $P(t+1)$ should be proved , so it goes
$ t^2 + 15t + 5≤21t^2 $
add $2t + 16$ to both sides to obtain $t^2 + 17t+21≤21t^2+2t+16$
From here it follows that $2t+16≤21t^2 $   $ ∀t∈Z^+$ So I can replace it with $42t^2+21$
i.e. $(t+1)^2+15(t+1)+5≤21(t+1)^2$
Is this way correct? If not can anyone suggest any other alternative?
 A: You've got the base case and the inductive hypothesis down pat.
What we assume to be true is called the inductive hypothesis: $$P(t): t^2 + 15t + 5≤21t^2\tag{P(t)}$$
Then, we need to prove that given $P(t)$, $P(t+1)$ is true, as you note:
$$P(t+1) = (t+1)^2 + 15(t+1) + 5 \leq 21(t+1)^2\tag{P(t+1)}$$
The objective is to manipulate the left hand side of $P(t)$, and to use the inductive hypothesis $P(t)$ to validate the inequality by showing the left-hand side is necessarily less than or equal to right-hand side of $P(t+1)$.
We can expand the left hand side of $P(t+1)$:
$$\begin{align}(t+1)^2 + 15(t+1) + 5 & = t^2 + 17t + 21 \\ \\ & = \color{blue}{\bf (t^2 + 15t + 5)} + (2t + 16) \\ \\ & \overset{\text{I.H.}}{\leq} \color{blue}{\bf 21t^2} + 2t + 16 \\ \\ &\leq 21t^2 + 42t + 21 \\ \\ &= 21(t^2 + 2 t + 1) \\ \\ &= 21(t+1)^2 \end{align}$$
...and end with the desired right-hand side of $P(t+1)$. Hence, we've proven $P(t) \implies P(t+1)$, and together with your base case, we're done!
A: Let $P(a)$ be the statement that $a^2+15a+5\le 21a^2$. Then certainly we have $P(1)$. Now suppose we have $P(t)$ for some positive integer  $t$, so that $t^2+15t+5\le 21t^2$.
$(t+1)^2+15(t+1)+5$
$=t^2+2t+1+15t+15+5$
$=t^2+15t+5+2t+16$
$\le 21t^2+2t+16$
$<21t^2+42t+21$
$=21(t+1)^2$
so we have $P(t+1)$.
