I want to prove that $x^2+y^2+xy\ge 0$ for all $x,y\in \mathbb{R}$.
My "proof": Suppose wlog that $x\ge y$, so $x^2\cdot x\ge x^2\cdot y\ge y^2\cdot y=y^3$ (because $x^2\ge 0$ so we can multiply both sides by it without changing the inequality sign) giving $x^3\ge y^3$. Substracting we have $x^3-y^3\ge 0$ or $(x-y)(x^2+xy+y^2)\ge 0$. Since $x\ge y$ we can divide by $x-y$ to get $x^2+xy+y^2\ge 0$.
Is it right?
Thanks for your help!
Note that we can prove it by $$x^2+y^2+xy=\left(x+\frac y2\right)^2+\frac{3}{4}y^2\ge 0.$$
The question posed was whether the OP's proof is correct. Mark Bennet pointed out one flaw in comments, namely that you can't divide by $x-y$ if $x=y$. It's worth noting another problem with the proof: You can indeed suppose wlog that $x\ge y$, from which $x^2\cdot x\ge x^2\cdot y$ follows (because $x^2\ge0$), but the full inequality, ending in $\ge y^2\cdot y=y^3$, does not. If, for example, $x=2$ and $y=-1$, we do not have $2^2\cdot(-1)\ge(-1)^2\cdot(-1)$.
One (simple) way:
let t = x2 + y2 + xy
then t - xy ≥ 0 since it is the sum of two real squares x2 + y2
and t + xy ≥ 0 since it is the square of the real (x + y) [since (x + y)2 = x2 + y2 + 2xy]
adding these, we get 2t ≥ 0, therefore t ≥ 0
Another approach: note that $-2|xy| \leq xy \leq 2|xy|$, so we have $$x^2 - 2|xy| + y^2 \leq x^2 + xy + y^2 \leq x^2 + 2|xy| + y^2$$ or equivalently, $$(|x|-|y|)^2 \leq x^2 + xy + y^2 \leq (|x| + |y|)^2$$ Since the left hand side is nonnegative, the result follows.
$(x+y)^2 \geq xy$
If one of $x, y$ is negative, this is obvious due to sign. Otherwise, WLOG, assume both $x$ and $y$ are positive. Since $x+y \geq x$ and $x+y \geq y$, then $(x+y)^2 \geq xy$.
My more logical, less formulaic proof: (its a bit long, but it covers all cases)
If x and y are both positive, all parts are positive
If one is positive and one is zero, all parts are positive, or zero (nothing negative)
If both are zero, all parts are zero, and it is not negative.
If both are negative, all of the parts are positive, as the signs cancel
If one is negative and one is zero, all parts are positive or zero (the square of the negative is positive, and the multiplied is zero)
The only case which is worrisome is when one is positive and one is negative
Let's say that x is positive, y is negative
This means that $x^2+y^2+xy$ is made of 3 parts, the first square is positive, the second square is positive, the third term, the multiplication is negative.
If $|x|>|y|$ then, multiplying both sides by x, we get $|x^2|>|xy|$ and we know the equation will be more positive than negative
If $|x|<|y|$ then, multiplying both sides by y, we get $|xy|<|y^2|$ and we know the equation will be more positive than negative
If $|x|=|y|$ then, multiplying both sides by x, we get $|x^2|=|xy|$ and we know the equation will be more positive than negative, because $y^2$ is positive and will make it positive, even though $x^2$ and $xy$ cancel out
A homogeneous bivariate second-degree polynomial with a negative discriminant can be only everywhere non-negative or everywhere non-positive over $\mathbb{R}^2$. Since in $(x,y)=(1,1)$ the inequality holds as $\geq$, for every $(x,y)\in\mathbb{R}^2$ we have $x^2+xy+y^2\geq 0$.
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ If $\quad\ds{x = y}$, $\quad\ds{x^{2} + y^{2} + xy = 3x^{2} \geq 0}$. It's also true whenever one of them is zero.
$\ds{\large\tt\mbox{Otherwise}}$:
In general, we can assume that $\ds{\verts{y} < \verts{x}}$: \begin{align} x^{2} + y^{2} + xy&={\pars{x^{2} + y^{2} + xy}\pars{x - y} \over x - y} ={x^{3} - y^{3} \over x - y} =x^{2}\,{1 - \pars{y/x}^{3} \over 1 - y/x} \end{align}
