How can I prove that $xy\leq x^2+y^2$ for all $x,y\in\mathbb{R}$ ?
$\endgroup$ 725 Answers
$\begingroup$$$x^2+y^2-xy=\frac{x^2}{2}+\frac{y^2}{2}+\frac{(x-y)^2}{2}$$
$\endgroup$ 3 $\begingroup$Use polar coordinates: $$x=r \cos \theta,y=r \sin \theta $$
Your inequality becomes $$r^2 \cos \theta \sin \theta \leq r^2 $$
which is pretty much trivial.
$\endgroup$ 9 $\begingroup$Though there are quite a few proofs already given, I'd like to add a visual one.
Another interesting answers - maybe the most interesting:
$$x^2+y^2-xy=\left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4} \geq 0.$$
$\endgroup$ 2 $\begingroup$\begin{align} 0\leq (x-y)^2 \implies & 0\leq x^2-2xy +y^2 \\ \implies & 2xy\leq x^2+y^2 \\ \implies & xy\leq \dfrac{x^2+y^2}{2} \\ \implies & xy\leq {x^2+y^2} \end{align}since the clearly nonnegative real $x^2+y^2$ clearly satisfies $\dfrac{x^2+y^2}{2} \leq x^2+y^2$.
$\endgroup$ 1 $\begingroup$A proof with only words: assume $x$ and $y$ are positive, if one is negative this is trivial. Then one of $x$ or $y$ is smallest, the other largest. Assume $y$ is the largest (else switch). Then $xy$ is clearly less than $y^2$, and adding $x^2$ doesn't change that!
$\endgroup$ 1 $\begingroup$For $xy<0$, this is trivial
You can do better by proving $2xy \leq x^2 + y^2$. Move $2xy$ to the right handside and factorise
$\endgroup$ 2 $\begingroup$$x^2+y^2=AC^2$
$\sin \angle BAC= \dfrac{BC}{AC}$
$\sin \angle BCA= \dfrac{AB}{AC}$
$\sin \angle BCA \cdot \sin \angle BAC <1 \implies \dfrac{xy}{x^2+y^2} <1$.
$\endgroup$ 0 $\begingroup$$$(x^2 - xy + y^2)(x + y) = x^3 + y^3.$$ $x^3 + y^3$ and $x + y$ have the same sign: $x \leq -y$ iff $x^3 \leq -y^3$. Therefore, $x^2 - xy + y^2 \geq 0$.
About the special case $x = -y$: it’s not really that special, but for a technical reason that goes beyond precalculus. Since non-trivial Zariski open sets are dense in the Euclidean topology, and polynomial maps are continuous between Euclidean topologies, all inequalities of the form $f(x_1, \ldots, x_n) \geq 0$ (where $f$ is a polynomial) that hold on a Zariski open set must hold everywhere, because $[0, +\infty)$ is closed and its preimage must thus contain the closure of said Zariski open set, i.e. the whole space. Therefore, such apparent special cases can be safely ignored as long as they’re Zariski closed and the inequality is not strict.
$\endgroup$ 2 $\begingroup$Another one - if $xy \le 0$ it is trivial, so let $xy > 0$. Then we have $$ 1 \le \dfrac{x}{y}+\dfrac{y}{x}$$
Now for any positive number, either it or its reciprocal must exceed $1$, unless both are $1$.
$\endgroup$ $\begingroup$If $xy$ is negative, then the statement is obvious since $xy < 0 \leq x^2 + y^2$.
Otherwise, $xy$ is non-negative, and we can show that $xy \leq 2xy \leq x^2+y^2$, where the latter follows from the trivial inequality $(x-y)^2 \geq 0 $.
$\endgroup$ $\begingroup$Just look at this picture below:
This is clearly a consequence of the concavity of the logarithm and the monotonicity of the exponential.
Since $xy\leq |x||y|$, we can assume that $x >0$ and $y > 0$. Then we have $$ xy = \exp(\ln(xy)) = \exp(\ln x + \ln y) = \exp(\frac12\ln x^2 + \frac12 \ln y^2) \leq \exp(\ln(\frac{x^2}2 + \frac{y^2}2)) =\frac{x^2}2 + \frac{y^2}2 \leq x^2 + y^2. $$ QED
$\endgroup$ 0 $\begingroup$Intuitive Approach
if $|x|\le |y|$ then we have $$x\cdot y \le y\cdot y \le y.y + x\cdot x$$ $$\Rightarrow x\cdot y \le x^2+y^2\tag1$$ Using Symmetry $(1)$ holds if $|x|\ge |y|$
$\endgroup$ 0 $\begingroup$$$x^2+y^2-xy=\frac{(2x-y)^2+3y^2}4=\frac{(2x-y)^2+(\sqrt3y)^2}4$$
Now, the square of any real numbers is $\ge0$
So, $(2x-y)^2+(\sqrt3y)^2\ge0,$ the equality occurs if each $=0$
$\endgroup$ 2 $\begingroup$Technique 1:$$\sqrt{x^2y^2} = xy \le \dfrac{x^2 + y^2}{2}\le x^2 + y^2$$ Technique 2:$$(x - y)^2 \ge 0 \implies x^2 +y^2 - 2xy \ge 0 \implies x^2 + y^2 \ge 2xy\ge xy$$ Technique 3 (my favorite): There's a statement $\dfrac{y}{x} + \dfrac{x}{y} \ge 2$ with many classical proofs (which I'd not state here). We can write the inequality as follows:$$\dfrac{y}{x}+\dfrac{x}{y} \ge 1$$Divide both sides by $xy$.$$\dfrac{1}{x^2} + \dfrac{1}{y^2} \ge \dfrac{1}{xy}$$Rewrite.$$\dfrac{x^2 + y^2}{x^2y^2} \ge \dfrac{xy}{x^2y^2}$$And finally...$$x^2 + y^2 \ge xy$$
$\endgroup$ $\begingroup$A different approach, $$A.M[x_i^n]\ge(A.M[x_i])^n\ge GM[x_1]^n$$
ie. AM of terms with $n$th power is greater than $n$th power of AM ie. greater than $n$th power of GM . So,$$\dfrac{x^2+y^2}2\ge(\dfrac{x+y}2)^2\ge(xy)$$ and so,$$(x^2+y^2)\ge xy$$
And for $xy<0$ the inequality is obvious.
$\endgroup$ 0 $\begingroup$Assume $x$ and $y$ are positive. Draw the right triangle with vertices $(0,0), (x,0), (x,y)$. Draw the circle around $(0,0)$ that circumscribes the triangle. Reflect the triangle in the $x$-axis. Reflect the two triangles in the $y$-axis too, for a better result.
$\endgroup$ $\begingroup$Let $y=kx$ then $ x xk \leq x^2 + x^2k^2$ since $k\leq k^2+1$ [in view of $0\leq (k-1/2)^2+3/4] $
And the ineq. clearly holds when $x=0$ . $\hspace {33mm} \blacksquare$
$\endgroup$ $\begingroup$$$(x-y)^2=x^2+y^2-2xy\\ (x-y)^2+2xy=x^2+y^2\\ 2xy\leq x^2+y^2$$ Therefore $xy\leq x^2+y^2$. Hence the proof.
$\endgroup$ $\begingroup$First, $$ a^2+3b^2\geq 0 $$ Now, $$ a^2+2ab+b^2+a^2-2ab+b^2\geq a^2-b^2 $$ Thus, $$ (a+b)^2+(a-b)^2\geq (a+b)(a-b) $$ Let $a+b=x, a-b=y$, i.e., $a=\frac{x+y}{2}, b=\frac{x-y}{2}$ which is one-to-one correspondence between $(x,y)$ and $(a,b)$, then, $$ x^2+y^2\geq xy $$ Q.E.D
$\endgroup$ $\begingroup$$(x - y)^2 \geq 0$
$x^2 + y^2 -2xy \geq 0$
$x^2 + y^2 \geq 2xy > xy$
upateHappy Green Kid is right below, the case for xy < 0 needs to be considered:
so, continuing from:
$x^2 + y^2 \geq 2xy$
if $xy > 0:$
$x^2 + y^2 \geq 2xy > xy$
else
$x^2 + y^2 > 0 > xy$
in both cases
$x^2 + y^2 > xy$
$\endgroup$ 1 $\begingroup$If xy is negative, then it is trivial. If both x and y are negative, then let x := -x and y := -y. This means that we can safely assume that both x and y are positive.
Square each sides yields
$$x^2y^2 \le x^4 + 2x^2y^2 + y^4$$
, which trivially holds.
$\endgroup$ $\begingroup$Say $y\ne 0$ then divide both sides $xy\le x^2+y^2$ by $y^2$; let $z=\frac{x}{y}$; then we need $z\le 1+z^2$ or equivalently $z^2-z+1\ge 0$; completing the square gives $(z-\frac{1}{2})^2+\frac{3}{4}\ge 0$ which is clear.
$\endgroup$ $\begingroup$Let $y>=x. xy<=x^2+y^2$. Divide both sides by $y$, and you get $x<=(x^2/y)+y$. If $x$ and $y$ are both positive numbers, this is clearly true because we said from the beginning that $y>=x$.
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