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Algebra Difficulty 4.6 AIME Prove it Hong Kong

Let aa, bb, cc, dd be positive real numbers such that a+b+c+d=1a + b + c + d = 1. Prove that
6(a3+b3+c3+d3)≥(a2+b2+c2+d2)+18. 6(a^3 + b^3 + c^3 + d^3) \ge (a^2 + b^2 + c^2 + d^2) + \frac{1}{8}.

Solution

We claim that f(x)=6x3−x2≥5x−18f(x) = 6x^3 - x^2 \ge \frac{5x-1}{8} for any x>0x > 0. Indeed,
6x3−x2≥5x−18⇔48x3−8x2−5x+1≥0⇔(4x−1)2(3x+1)≥0. \begin{align*} & 6x^3 - x^2 \ge \frac{5x-1}{8} \\ \Leftrightarrow \quad & 48x^3 - 8x^2 - 5x + 1 \ge 0 \\ \Leftrightarrow \quad & (4x-1)^2(3x+1) \ge 0. \end{align*}
This clearly holds. It follows that
f(a)+f(b)+f(c)+f(d)≥5(a+b+c+d)−48=18. f(a) + f(b) + f(c) + f(d) \ge \frac{5(a+b+c+d) - 4}{8} = \frac{1}{8}.
Equality holds when a=b=c=d=14a = b = c = d = \frac{1}{4}.

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