Maths Olympiad Prep

Track / Stage 4 / 58 of 340 #318 of 1964

Problem 318

AMC 12 late, AIME early
Algebra Difficulty 4.6 Prove it India — Team Selection Test · India · 2008

Let aa, bb, cc be positive real numbers such that a2+b2+c2<2(a+b+c)a^2 + b^2 + c^2 < 2(a + b + c). Prove that
3abc<4(a+b+c). 3abc < 4(a + b + c).

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

The Cauchy-Schwarz inequality gives
(a+b+c)2≤3(a2+b2+c2). (a + b + c)^2 \le 3(a^2 + b^2 + c^2).
Thus (a+b+c)<6(a + b + c) < 6 and hence (a+b+c)39<4(a+b+c)\frac{(a + b + c)^3}{9} < 4(a + b + c). Now the AM-GM inequality gives
(a+b+c)3≥27abc. (a + b + c)^3 \ge 27abc.
Thus
3abc≤(a+b+c)39<4(a+b+c). 3abc \le \frac{(a + b + c)^3}{9} < 4(a + b + c).

Alternately, we have a+b+c<6a + b + c < 6 so that a2+b2+c2<2(a+b+c)<12a^2 + b^2 + c^2 < 2(a + b + c) < 12. Thus
(a2+b2+c2)(a+b+c)<12(a+b+c). (a^2 + b^2 + c^2)(a + b + c) < 12(a + b + c).
Using the AM-GM inequality, we have
a2+b2+c2≥3(abc)2/3,a+b+c≥3(abc)1/3. a^2 + b^2 + c^2 \ge 3(abc)^{2/3}, \quad a + b + c \ge 3(abc)^{1/3}.
Thus
9abc≤(a2+b2+c2)(a+b+c)<12(a+b+c), 9abc \le (a^2 + b^2 + c^2)(a + b + c) < 12(a + b + c),
giving the required inequality. Thus

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