This is what I am asking. I do not want you to prove these inequalities for me; instead, I would like you to teach me a general method with which I can prove the inequality:
$$x^2 + xy + y^2 \ge 0$$
If you could prove a similar inequality to illustrate the general method, I will be thankful.
This is a homogeneous polynomial in $x$ and $y$. So set, e.g., $y=tx$, and rewrite it as $$x^2+xy+y^2=x^2(1+t+t^2)$$ So all you have to prove is that $1+t+t^2\ge 0$.
If $x=y$, then $x^2+xy+y^2=3x^2\ge 0$.
If $x>y$. Then: $$x^3> y^3 \iff x^3-y^3> 0 \iff (x-y)(x^2+xy+y^2)> 0 \iff x^2+xy+y^2> 0.$$
Can you check the case $x<y$?
Your first inequality is $$x^2+2x\frac{y}{2}+\frac{y^2}{4}+y^2-\frac{y^2}{4}\geq 0$$ (completing the square) For the second inequality $$x$$ is not a solution ,then we can write $$x^2+\frac{1}{x^2}-3(x+\frac{1}{x})+4=0$$ then Substitute $$t=x+\frac{1}{x}$$
Assuming $x,y$ to be real,
$x^2+xy+y^2=\dfrac{(x+2y)^2+3y^2}4$
If $z/x^2=(x+1/x)^2-3(x+1/x)+4-2=(x+1/x-1)(x+1/x-2)$
Now $x+1/x\ge2$ or $\le-2$
Any $x^2+cxy+y^2$ with $-2\le c\le 2$ is a convex combination of $x^2-2xy+y^2=(x-y)^2\ge 0$ and $x^2+2xy+y^2=(x-y)^2\ge 0$, hence non-negative.
Here is another way:
Show $x^2+xy+y^2 \ge 0$.
1) If $x,y\ge 0:$
$x^2,y^2, xy \ge 0$, the sum is $\ge 0$√.
2) If $x,y \le 0:$
$x^2,y^2, xy \ge 0$, as before the sum $\ge 0$√.
3) Let $xy \le 0$.
Assume $x \lt 0$, $y \ge 0$.
Consider $m:=\max(|x|, y).$
$x^2+y^2+xy = x^2+y^2-|x|y \ge$
$ x^2+y^2-m^2 \ge 0$ (Why?).
Assume $y \lt 0$, $x \ge 0$, and complete the proof.
Since
$$\begin{equation}\begin{split} y^{2}+ yx+ x^{2}= (- y+ x)^{2}+ 3yx= (y+ x)^{2}- yx \end{split}\end{equation}$$
Since
$$\begin{equation}\begin{split} y^{2}+ yx+ x^{2}= \frac{y^{3}- x^{3}}{y- x}\geqq 0\because \lim_{y\rightarrow x+ 0^{+}}\frac{y^{3}- x^{3}}{y- x}= 3x^{2}\geqq 0, \lim_{y\rightarrow x+ 0^{-}}\frac{y^{3}- x^{3}}{y- x}= 3x^{2}\geqq 0 \end{split}\end{equation}$$
