Acapulco, February 25 – In his first competitive match since his defeat in a thrilling Australian Open semi-final against Carlos Alcaraz, world No. 4 Alexander Zverev advanced to the second round of the Mexican Open with a 6-2, 6-4 victory over Frenchman Corentin Moutet.
With his 15th win in Acapulco, Zverev (15-5) tied Rafael Nadal (15-2) for the most wins at the event since it switched to hard courts in 2014. He also surpassed 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 won the first set in under 40 minutes, thanks to his serve, which racked up five aces and broke his opponent's serve twice. In the second set, Moutet rallied, but could not compete 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 is not easy, but when I had the opportunity, I won the match. Acapulco is not an easy place to play because of the conditions, but I feel I am 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 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) to Chinese Yibing Wu (No. 142), who came from the qualifying round.
Wu's next opponent will be Japan's Sho Shimabukuro, who eliminated Frenchman Adrian Mannarino.
