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Algebra Difficulty 4.5 AIME Prove it Croatia

Prove that for all real numbers a,b,ca, b, c the following inequality holds
13(a+b+c)2≤a2+b2+c2+2(a−b+1). \frac{1}{3}(a+b+c)^2 \leq a^2 + b^2 + c^2 + 2(a-b+1).

Solution

Transforming the right hand side of the inequality we get
a2+b2+c2+2(a−b+1)=a2+2a+1+b2−2b+1+c2=(a+1)2+(b−1)2+c2. \begin{aligned} a^2 + b^2 + c^2 + 2(a - b + 1) &= a^2 + 2a + 1 + b^2 - 2b + 1 + c^2 \\ &= (a + 1)^2 + (b - 1)^2 + c^2. \end{aligned}
From the A–K inequality it follows
(a+1)2+(b−1)2+c23≥(a+1)+(b−1)+c3 \sqrt{\frac{(a+1)^2 + (b-1)^2 + c^2}{3}} \ge \frac{(a+1) + (b-1) + c}{3}
i.e.
(a+1)2+(b−1)2+c2≥13(a+b+c)2. (a + 1)^2 + (b - 1)^2 + c^2 \ge \frac{1}{3} (a + b + c)^2.

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