Showing $4x + 4b^2 x \leq 4 + (x + b^2 x)^2$

Question

I am working on a problem, and I got to a point where I will be done if I can show that for any $0 \leq x \leq 2$ and any constant $b$, $$ 4x + 4b^2 x \leq 4 + (x + b^2 x)^2. $$

As a sanity check, I've checked that the equation holds for $x = 0, x = 1,$ and $x = 2$. How can I extend this to other all $x$ between $0$ and $2$?

Answer 1

Let $y=b^2x$ then

$$ 4x + 4y \leq 4 + (x + y)^2. $$ and let $t=x+y$ $$ 4t \leq 4 + t^2. $$ which is true since: $0\leq (t-2)^2$.

Answer 2

Guide: It holds true for more $x$ than what you expected.

$$4(x+b^2x) \le 4 + (x+b^2x)^2$$

$$(x+b^2x)^2-2(2)(x+b^2x)+2^2 \ge 0$$

Can you use the formula $(a-b)^2=a^2-2ab+b^2$ to simplify the inequality above?