The Best of the Day preview looks for a forehand winner with tennis tips for Alex de Minaur’s Italian Open assignment against Marco Cecchinato.
22:00 (AEST) @ Foro Italico, Rome
History: First meeting
Final Thoughts: Aussie wunderkind Alex de Minaur heads into his first-up assignment at the Italian Open as a rank outsider.
— Internazionali Bnl (@InteBNLdItalia) May 10, 2019
World No.19 Marco Cecchinato is only five ranking places ahead of the 20-year-old de Minaur, but the Italian is a clay-court specialist.
Cecchinato boasts a stack of Challenger and Futures titles on the surface, while his three ATP titles were all won on clay in the past 13 months. Cecchinato beat John Millman in the final of the Hungarian Open last April, while he won the Croatia Open Umag last year and the Argentina Open in February.
Meanwhile, he came from nowhere to reach the French Open semis last year, eliminating Novak Djokovic. Cecchinato had never previously won a grand slam match.
But he comes into this tournament off a first-round Madrid Open loss to Diego Schwartzman, who he defeated in the final of the Argentina Open.
De Minaur does have a couple of Futures titles on clay to his name. He announced himself as one of the genuine rising stars of men’s tennis with his maiden ATP triumph in January, winning the Sydney International.
He then reached the third round of the Australian Open – his third straight appearance in the last 32 of a grand slam.
But the injury-hampered tyro comes into this tournament in underwhelming form. Since reaching the quarters of the Mexican Open in February, he has bombed out in the first round of the Indian Wells Masters, the Estoril Open (as the sixth seed) and the Madrid Open.
Australian No.1 Alex de Minaur's return to tennis from injury has failed to gather any momentum with a straight-sets loss in the first round of the Madrid Open.
— The Age Sport (@theagesport) May 8, 2019
An upset win may be beyond $3.81 underdog de Minaur. But he is a dogged competitor and will be determined to make Cecchinato scrap for every point.