Solving inequality $\frac{7x+12}{x} \geq 3$ I was sitting with my friend in a park, when he thought of and asked for solving this inequality:
$$
\frac{7x+12}{x} \geq 3
$$
I was confident enough that I could solve this example as continued:
$$
7x + 12 \geq 3x
$$
$$
4x \geq -12
$$
$$
x \geq -3
$$
He simply told me to put $x = -4$, and in the equation, I applied:
$$
\frac{-28+12}{-4}
$$
It got $\frac{-16}{-4}$ which is 4, so it is false that $x \geq -3$. Where I am doing a mistake, and what would be the correct solution.

You made a mistake when you multiplied both sides of the inequality by $x$.  Since multiplying an inequality by a negative number reverses the direction of the inequality, you have to consider cases when you multiply both sides of the inequality by $x$.
Method 1
Case 1:  $x > 0$
\begin{align*}
\frac{7x + 12}{x} & \geq 3\\ 
7x + 12 & \geq 3x\\
4x + 12 & \geq 0\\
4x & \geq -12\\
x & \geq -3
\end{align*}
Since $x > 0$ and $x \geq -3$, $x > 0$.
Case 2:  $x < 0$
\begin{align*}
\frac{7x + 12}{x} & \geq 3\\
7x + 12 & \leq 3x\\
4x + 12 & \leq 0\\
4x & \leq -12\\
x & \leq -3
\end{align*}
Since $x < 0$ and $x \leq -3$, $x \leq -3$.
Since the two cases are mutually exclusive and exhaustive, $x > 0$ or $x \leq -3$.  Therefore, the solution set is
$$S = (-\infty, -3] \cup (0, \infty) = ]-\infty, -3] \cup ]0, \infty[$$
Method 2:
We can avoid cases if we first subtract $3$ from each side of the inequality.
\begin{align*}
\frac{7x + 12}{x} & \geq 3\\
\frac{7x + 12}{x} - 3 & \geq 0\\
\frac{7x + 12}{x} - \frac{3x}{x} & \geq 0\\
\frac{4x + 12}{x} & \geq 0\\
\frac{x + 3}{x} & \geq 0
\end{align*}
Equality holds when $x = -3$.  The strict inequality holds when $x + 3$ and $x$ have the same sign.  $x + 3 > 0$ and $x > 0 \implies x > 0$.  $x + 3 < 0$ and $x < 0 \implies x < -3$.  Thus, the solution set is
$$S = \{-3\} \cup (-\infty, -3) \cup (0, \infty) = (-\infty, -3] \cup (0, \infty)$$
Method 3
Since the direction of the inequality is preserved if we multiply both sides of the inequality by a positive number, we multiply both sides of the inequality by $x^2 > 0$.
\begin{align*}
\frac{7x + 12}{x} & \geq 3\\
x(7x + 12) & \geq 3x^2\\
7x^2 + 12x & \geq 3x^2\\
4x^2 + 12x & \geq 0\\
x^2 + 3x & \geq 0\\
x(x + 3) & \geq 0
\end{align*}
Equality holds if $x = 0$ or $x = -3$.  However, the original expression is not defined when $x = 0$, so equality holds if and only if $x = -3$.  The strict inequality holds when $x + 3$ and $x$ have the same sign.  $x + 3 > 0$ and $x > 0 \implies x > 0$.  $x + 3 < 0$ and $x < 0 \implies x < -3$.  Thus, the solution set is
$$S = \{-3\} \cup (-\infty, -3) \cup (0, \infty) = (-\infty, -3] \cup (0, \infty)$$
