we have $a_2^2 + \cdots +a_{100}^2 \geq \frac{(a_2 + \cdots +a_{100})^2}{99}$ so $101=a_1^2+ a_2^2 + \cdots +a_{100}^2 + ( a_1+a_2 + \cdots + a_{100})^2 \geq 2a^2_1+\frac{(a_2 + \cdots +a_{100})^2}{99}+(a_2 + \cdots +a_{100})^2+2a_1(a_2 + \cdots +a_{100})$ consider it for a polynomial whose degree is 2. we will easily get,

This is inequality was proposed by Dinu Serbanescu at jbmo test selection in 2004.

by cauchy: $(1+1+\cdots + 1)(a_2^2+a_3^2 +\cdots a_{100}^2)\ge(a_2 + a_3\cdots +a_{100}^2)$ so $a_2^2 + \cdots +a_{100}^2 \geq \frac{(a_2 + \cdots +a_{100})^2}{99}$ As in post #2, we have $101=a_1^2+ a_2^2 + \cdots +a_{100}^2 + ( a_1+a_2 + \cdots + a_{100})^2 \geq 2a^2_1+\frac{(a_2 + \cdots +a_{100})^2}{99}+(a_2 + \cdots +a_{100})^2+2a_1(a_2 + \cdots +a_{100})$ Setting $a_1^2=x$ and $(a_2 + \cdots +a_{100})^2=y$, we have $2x^2+\frac{100}{99}y^2+2xy=101$, and the max value for $a_1$ can be found by lagrange multipliers or quadratic discriminant. Finally, repeat this logic for all $a_n$