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by IANS |
Acapulco, Feb 25 (IANS) In his first competitive action since his defeat in an epic Australian Open semi-final against Carlos Alcaraz, world no. 4 Alexander Zverev advanced to the second round of the Mexcian Open with a 6-2,6-4 win against Frenchman Corentin Moutet.
With his 15th win in Acapulco, Zverev (15-5) moved level with Rafael Nadal (15-2) for most wins at the event since it switched to hard courts in 2014. He also passed David Ferrer for the second-most ATP 500 wins since the series began in 2009, with his 117 victories (117-49) trailing only Nadal's 121 (121-19).
Zverev took the first set in under 40 minutes, thanks to his serve racking up five aces and breaking his opponent's serve twice. In the second set, Moutet rallied but could do little against Zverev's dominant performance, particularly his two-handed backhand and consistent net play.
This was the third time the German and the Frenchman had faced each other on the ATP circuit, with a positive record for "Sasha", who had beaten him twice in 2025.
“It wasn’t easy (the match), he hit several drop shots and that style of tennis isn’t easy, but when I had the opportunity I won the match. Acapulco isn’t an easy place to play because of the conditions, but I feel I’m playing at a good level,” Zverev said in the post-match press conference.
Seeking to repeat his title triumph at the ATP 500 in 2021, Zverev's will face Serbian Miomir Kecmanovic, whom the German has faced three times, with two wins for the German.
Earlier, the runner-up of the tournament in 2024 Casper Ruud was eliminated after losing 6-7 (2-7), 6-7 (2-7) against the Chinese Yibing Wu (No. 142), who comes from the qualifying round.
Wu's next opponent will be Japan's Sho Shimabukuro, who eliminated Frenchman Adrian Mannarino.
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