Setting up inequalities is proving difficult Given $$(a+b)^2\leq4$$
$$(a-b)^2\leq4$$
Is this true?
$$2(a+b)^2+2(a-b)^2-b^2\leq16$$
The problem I need to solve is to find the maximum value of $(4a^2+3b^2)$ Please help.

$(a+b)^2 \le 4$ so $2(a+b)^2 \le 8$.
$(a-b)^2 \le 4$ so $2(a-b)^2 \le 8$.
So $2(a+b)^2 + 2(a+b)^2 \le 8+8 = 16$.
And $b^2 \ge 0$ so
$2(a+b)^2 + 2(a+b)^2-b^2 \le 2(a+b)^2 + 2(a+b)^2 \le 16$.
$16 \ge 2(a+b)^2 + 2(a+b)^2-b^2 = 2(a^2 + 2ab +b^2) + 2(a^2 -2ab + b^2) -b^2 = 4a^2 + 3b^2$.  So $16$ is a potential maximum.  But we don't know if it is an achievable maximum.
But if $(a+b)^2 = 4$ its maximum; and $(a -b)^2=4$ its maximum; and $b^2 = 0$; it's minimum then $4a^2 + 3b^2$ will meet its maximum of $16$.
$b^2 = 0 \implies b=0$ and so $(a+b)^2 = a^2 = (a-b)^2 =4$ which implies $a=\pm 2$ allow $4a^2 + 3b^2$ to equal a maximum possible value of $16$
