WBSS: Ryan Burnett ready to prove he is world's best bantamweight
Tuesday 19 June 2018 15:30, UK
Ryan Burnett has vowed to prove he is the world's number one bantamweight by winning the World Boxing Super Series.
The WBA 'Super' champion is part of a star-studded line-up - featuring Emmanuel Rodriguez (IBF), Zolani Tete (WBO) and Naoya Inoue (WBA 'regular') - which will compete for the coveted Muhammad Ali trophy.
Burnett, who hopes to kick-off the knockout competition at home in Belfast, believes he is only just coming into his prime and is confident of showing his worth by winning the Series.
The fans want to see the best against the best and that's exactly what will happen in this tournament.
When asked whether the tournament will decide who the world's best bantamweight is, Burnett, speaking exclusively to Sky Sports, astutely replied: "Yes and it will be me.
"This is a massive tournament and I'm blessed to be a part of it. I'm very confident of winning the whole thing, I go into every fight very confident.
Emmanuel Rodriguez ready to prove he is world's best bantamweight
"I have an incredible team around me and I'm just starting to step into my prime. I've just turned 26 so things are good.
"I definitely want to have the first fight in Belfast. I put bums on seats and everyone in the city will be buzzing for such a massive tournament, so I'll be pushing to get my opener back home.
"It's incredible really, I grew up watching fighters who inspired me and wanted to be like them. Now, I'm up there and leading the way so to speak."
Although he acknowledges that the latest instalment of the Super Series is filled with "incredible fighters", Burnett is unfazed and has no preference on who he could take on.
"The fans want to see the best vs the best and that's exactly what will happen in this tournament," he said.
"I haven't really thought about who I'll face, all the lads in the tournament are incredible fighters, so any of them.
"I don't really have one I've picked that I want to face, I'm just confident of beating whoever."
