Let $a,b$ be real numbers. Suppose that inequality $|(ax+by)(ay+bx)|\leqslant x^2+y^2$ holds for any real $x,y$. Prove that $a^2+b^2\leqslant2$

Question

I know that this question has already been asked, but date of sending solutions passed (30th Sep), so this part of contest is inactive.

$$|(ax+by)(ay+bx)|\le x^2+y^2$$ $$|ab(x^2+y^2)+xy(a^2+b^2)| \le x^2+y^2$$ $$|k(a^2+b^2)+ab| \le 1$$ where $k=\frac{xy}{x^2+y^2}$

$-\frac12\le k \le \frac12$ since $x^2+y^2\ge \pm 2xy$

Then I tried to treat left hand side as quadratic, but it did not lead to anything useful. On the other hand from https://www.desmos.com/calculator/ljmcqbpobn it follows that there are $a,b$ s.t $a^2+b^2>2$, so I guess there are some mistakes in my approach.

Answer 1

For $x=y=1$: $(a+b)^2\le 2$, so $a^2+b^2+2ab\le 2$.

For $x=1$ and $y=-1$: $(a-b)^2\le 2$, so $a^2+b^2-2ab\le 2$.

$$a^2+b^2=\frac{(a^2+b^2+2ab)+(a^2+b^2-2ab)}{2}\le\frac{2+2}{2}=2.$$

Answer 2

(Using your approach) You're real close.

Show that if $ a^2 + b^2 > 2$, then we can find some $ k$ in that range such that $|k (a^2 + b^2) + ab| > 1$, which is a contradiction. Thus $ a^2 + b^2 \leq 2$.

Hint:

It might be helpful to separately consider cases of $ ab \geq 0, ab < 0$.

Answer 3

I will write a separate answer because I think you misunderstood the problem and therefore cannot benefit from the solutions.

You don't have to prove anything for all $x,y$. Instead you are told in advance that the inequality holds for all $x,y$ and you have to deduce something about $a,b$. If you can deduce all you need by only looking at two pairs of $x,y$ then, so to speak, you already win.

To give a hugely oversimplified example, consider the following problem:

Suppose that for all $x$ we have $ax^2\geq 0$. Prove that $a\geq 0$.

Of course you can make some argument that takes into account all values of $x$, but it is easier to say: if it holds for all $x$ then it holds in particular for $x=1$, and therefore $a\cdot 1\geq 0$, so we are done. Someone else could come and ask you also to plug in $x=2$ and $x=3$ but all you would get is $4a\geq 0$ and $9a\geq 0$, which you already know.

It is the same with the OM problem. If you can prove all you need by testing only two pairs $(x,y)$, good for you.