<s>[INST] $3x^2+y^2 \ge -ax(x+y)$, solve for $a$ so this inequality is true for all $x,y$ I started trying to reorganize the equation
$$3x^2+y^2\ge -ax(x+y)$$
$$3x^2+y^2+ax^2+axy\ge0$$
notice:$$\left(\frac {ax}2+y\right)^2=axy+\frac14x^2a^2+y^2$$
now i should be able to simply add $\frac14x^2a^2$ as this number is always bigger or equal to zero so it wont affect the inequality?
$$x^2(3+a)+\left(\frac {ax}2+y\right)^2\ge0$$
this inequality should hold for all $x,y$ when $3+a \ge 0, a\ge-3$
but this is wrong, where is the mistake?
 [/INST] We have that
$$3x^2+y^2\ge-ax(x+y) \iff(3+a)x^2+axy+y^2 \ge 0$$
which requires $a\ge -3$ for $y=0$ and for $y\neq 0$ by $\frac x y =t$ we have
$$\iff y^2\left((3+a)t^2+at+1\right) \ge 0 \iff a^2-4(3+a) \le 0 \iff a^2-4a-12 \le 0$$
which requires $-2\le a \le 6$.
Note that adding the factor $\frac14x^2a^2$ affects the inequality, for $x\neq 0$, for that reason your solution can't be right.
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