|(ax+by)(ay+bx)| < x^2+y^2

[Guest]

If [tex]|(ax+by)(ay+bx)|\le x^2+y^2[/tex] holds for all real x and y, then prove [tex]a^2+b^2\le 2[/tex]

[shyamjayakannan]

Put [tex]x=y=1\Rightarrow|(a+b)(a+b)|\le2\Rightarrow a^2+b^2+2ab\le2...(1)[/tex]
Put [tex]x=1,\ y=-1\Rightarrow|-(a-b)(a-b)|\le0\Rightarrow a^2+b^2-2ab\le0\Rightarrow a^2+b^2\le2ab...(2)[/tex]

From (2), [tex]a^2+b^2\le2ab\Rightarrow2\left(a^2+b^2\right)\le a^2+b^2+2ab\le2[/tex] [From (1)]. So, [tex]2\left(a^2+b^2\right)\le2\Rightarrow a^2+b^2\le1[/tex]

Since [tex]a^2+b^2\le1\Rightarrow\boxed{a^2+b^2\le2}[/tex]