Direct proportion - sum of numbers and square root of sum of the squared numbers

Question

Is it true for all the cases that if $x + y > a+b$, then $\sqrt{x^2 + y^2} > \sqrt{a^2 + b^2}$? In other words - is there a direct proportion between sum of numbers and square root of the sum of those numbers squared?

Answer 1

No, take $x=y=3,a=5,b=0$, then $x+y=6, a+b=5, \sqrt{x^2+y^2}=\sqrt {18} =3\sqrt 2\approx 4.242, \sqrt{a^2+b^2}=5$

Answer 2

A more easy example to notice is if $x=y=1$ and $a=b=-2.$ Then you have: $$2=x+y\lt a+b=-4\quad\color{grey}{\text{but}}\quad \sqrt{x^2+y^2}=\sqrt2<\sqrt{(-2)^2+(-2)^2}=\sqrt{8}.$$