Reverse Triangle Inequality inner product

Question

I have a problem proving that for an inner product space over $\mathbb{F}$ this happens: $$ \|v+u\|\ge\bigl|\|v\|-\|u\|\bigr| $$ My attempt: $$ \|v+u\|=\sqrt{\langle v,u\rangle+\langle u,v\rangle+\langle v,v\rangle+\langle u,u\rangle} $$ $$ \bigl|\|v\|-\|u\|\bigr|=\left|\sqrt{\langle v,v\rangle}-\sqrt{\langle u,u\rangle}\right| $$ after squaring both sides of the equation we get $$ \langle v,u\rangle+\langle u,v\rangle+\langle v,v\rangle+\langle u,u\rangle?\langle v,v\rangle-2\sqrt{\langle v,v\rangle\langle u,u\rangle}+\langle u,u\rangle $$ we can cancel the same elements and we stay with: $$ \langle u,v\rangle+\langle v,u\rangle?-2\sqrt{\langle v,v\rangle\langle u,u\rangle} $$ after using $\langle v,u\rangle = \overline{\langle u,v\rangle}$ we stay with: $$ 2\operatorname{Re}\langle u,v\rangle ? -2\sqrt{\langle v,v\rangle \langle u,u\rangle } $$ After squaring yet again I stay with $$ \bigl(\operatorname{Re}\langle u,v\rangle\bigr)^2 ?\langle v,v\rangle \langle u,u\rangle $$ How do I go on to prove anything from this? Is my approach completely wrong?

Answer 1

The approach indicated by @Bungo is doable

let $u = (u+v)-v$

through triangle inequality we get $\|v\| \leq \|u+v\| + \|-u\| = \|u+v\| + \|u\|$ for $\|-u\|=\|u\|$

that is $\|v\|\leq\|v+u\|+\|u\|$

again by letting $v=(v+u)-u$

similarly we get $\|u\|\leq\|v+u\|+\|v\|$

therefore $\|v+u\|$ is greater than or equal to both $\|v\|-\|u\| and -(\|v\|-\|u\|)$

thus $\|v+u\|\geq|\|v\|-\|u\||$