Do not lose the condition we can assume that: $0<x_1\leq x_2\leq\cdots\leq x_9.$ So$ (x_7+x_8+x_9)^2\geq x_1^2+x_2^2+x_3^2+\cdots+x_9^2=25.$ So $x_7+x_8+x_9\geq 5$ (DONE.)

the sum of the three largest terms is minimilised when all 9 are equal. let them all be x. then 9x^2=25 , x=5

By cauchy schwarz we have: (x1^2 + x2^2 + ... + x9^2)(1 + 1 + ... + 1) ≥ (x1 + x2 + ... + x9)^2 25 * 9 ≥ (x1 + x2 + ... + x9)^2 ---> 15 ≥ x1 + x2 + ... + x9 a = x1 + x2 + x3 b = x4 + x5 + x6 c = x7 + x8 + x9 15 ≥ a + b + c ---> one of a,b,c is at least 5

Mahdi_Mashayekhi wrote: (...) 15 ≥ a + b + c ---> one of a,b,c is at least 5 I don't think that follows, it rather follows that at least one of a, b, c is at most 5