/sci/ - Science & Math

>>11908943
It is enough to deal with [math]||y|| \geq ||x||[/math], since otherwise leftside <= 1 <= rightside.
[math]||x+y|| \geq ||y|| - ||x|| [/math] so it is enough to prove
[eqn]\frac{1+||y||^2}{1 + ||x||^2} \leq (1 + ||y|| - ||x||)^2. [/eqn]
Multiply sides by [math]1+||x||^2[/math], move everything on one side, and for simpler notation put [math]a=||x||[/math], [math]b = ||y||[/math].
You will arrive at some ugly polynomial expression involving a, b. The original inequality is an equality when [math]||x||=||y||[/math] and this tells you that you can factor out (a-b) from this ugly expression. Ultimately you arrive at
[eqn](b - a) (-a^3 + a^2 b + 2 a^2 - 2 a + 2) \geq 0[/eqn]
which is clearly true when b>=a.

>>11909351
good morning!