Prove that for real numbers $a,b,c,d,e$ which belong to the interval $[0, 1]$ this always holds: $$(1+a+b+c+d+e)^2 \ge 4(a^2+b^2+c^2+d^2+e^2)$$ When are the two sides of inequality equal?
2026-03-30 12:02:02.1774872122
Algebra: Inequality in interval $[0, 1]$ (Junior Serbian Mathematical Olympiad 2014)
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hint: $0\le x \le 1 \implies 0\le x^2 \le x \le 1, (1+s)^2 \ge 4s, s = a^2 + b^2 + c^2 + d^2 + e^2$. Let $x = a, b, c, d, e$ and add up.