Knowing that $x^2+y^2 = 2 $ prove that $x+y \le 2$

I rewrite this as $$x^2+y^2 \ge x+y$$ Now, multiply both sides by 2 $$x^2 -2x + 1 + y^2-2y + 1 -2 +x^2 + y^2 \ge 0$$ I substitue $2$ for $x^2+y^2$ $$(x-1)^2+(y-1)^2\ge0$$ Which is true

  • $\begingroup$ Now it's true! I fixed your post. $\endgroup$ – Michael Rozenberg May 7 '17 at 20:20
  • 1
    $\begingroup$ I don't understand how you obtain the second line multiplying everything by $2$. $\endgroup$ – Bernard May 7 '17 at 20:22
  • $\begingroup$ @Bernard $x^2 + x^2 + y^2 + y^2 - 2x - 2y + 1 + 1 - 2 \ge 0$ Now I group the terms to get this expression. $\endgroup$ – ILoveChess May 7 '17 at 20:39
  • 2
    $\begingroup$ Um, your very first line you assume what you want to proof and then you ... don't get a contradiction. That's not a proof. So what if you get something is true? If polar bears lived in the middle of the Indian Ocean then all they would have to eat is fish. So polar bears would eat fish. Which is true; polar bear do eat fish. Therefore polar bears live in the Indian Ocean... because they eat fish... which they'd have to do if they lived in the Indian Ocean... so they do.... $\endgroup$ – fleablood May 8 '17 at 6:22
  • $\begingroup$ An alternative to Hagen's proof below is to use differentiability: every point on the circle is $(x,y)=(\sqrt2\cos t,\sqrt2\sin t)$ for some $t$ hence one must show that $f(t)\leqslant\sqrt2$, where $f(t)=\cos t+\sin t$. The derivative of $f$ is such that $f'(t)=\cos t-\sin t$ hence $f(t)$ is maximum at $t=\pi/4$ and $f(\pi/4)=\sqrt2$, qed. $\endgroup$ – Did May 10 '17 at 5:40

Your proof is insecure and accidentally. When a line $x+y=t$ is tangent to circle, t is max or minimum.


Or (shortened, once we know how to poceed): $$x+y\le x+y+\tfrac{ (x-1)^2+(y-1)^2}2=x+y+\tfrac{x^2-2x+1+y^2-2y+1}2=\frac{x^2+y^2+2}2=2$$


Using Cauchy-Schwarz:

$$(1+1)(x^2+y^2)\ge(x+y)^2 \Rightarrow (x+y)^2\le4 \Rightarrow |x+y|\le2.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.