How can I prove that $xy\leq x^2+y^2$? [closed]

Question

How can I prove that $xy\leq x^2+y^2$ for all $x,y\in\mathbb{R}$ ?

Answer 1

$$x^2+y^2-xy=\frac{x^2}{2}+\frac{y^2}{2}+\frac{(x-y)^2}{2}$$

Answer 2

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.

Answer 3

Though there are quite a few proofs already given, I'd like to add a visual one.

Answer 4

\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 $\dfrac{x^2+y^2}{2} \leq x^2+y^2$.

Answer 5

Another interesting answers - maybe the most interesting:

$$x^2+y^2-xy=\left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4} \geq 0.$$

Answer 6

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!

Answer 7

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

Answer 8

$x^2+y^2=AC^2$

$\sin \angle BAC= \dfrac{BC}{AC}$

$\sin \angle BCA= \dfrac{AB}{AC}$

$\sin \angle BCA \cdot \sin \angle BAC <1 \implies \dfrac{xy}{x^2+y^2} <1$.

Answer 9

$$(x^2 - xy + y^2)(x + y) = x^3 + y^3.$$ $x^3 + y^3$ and $x + y$ have the same sign: $x \leq -y$ iff $x^3 \leq -y^3$. Therefore, $x^2 - xy + y^2 \geq 0$.

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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.

Answer 10

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 $1$.

Answer 11

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 $.

Answer 12

Just look at this picture below:

Answer 13

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

Answer 14

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 $|x|\ge |y|$

Answer 15

$$x^2+y^2-xy=\frac{(2x-y)^2+3y^2}4=\frac{(2x-y)^2+(\sqrt3y)^2}4$$

Now, the square of any real numbers is $\ge0$

So, $(2x-y)^2+(\sqrt3y)^2\ge0,$ the equality occurs if each $=0$

Answer 16

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...$$x^2 + y^2 \ge xy$$

Answer 17

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.

Answer 18

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.

Answer 19

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 $x=0$ . $\hspace {33mm} \blacksquare$

Answer 20

$$(x-y)^2=x^2+y^2-2xy\\ (x-y)^2+2xy=x^2+y^2\\ 2xy\leq x^2+y^2$$ Therefore $xy\leq x^2+y^2$. Hence the proof.

Answer 21

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

Answer 22

$(x - y)^2 \geq 0$

$x^2 + y^2 -2xy \geq 0$

$x^2 + y^2 \geq 2xy > xy$

upate Happy 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

$x^2 + y^2 > xy$

Answer 23

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.

Answer 24

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.

Answer 25

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 $y>=x$.