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Algebra Difficulty 4.7 AIME Prove it Greece

Let aa, bb, cc, dd be positive real numbers such that
a2+b2+c2+d2=4. a^2 + b^2 + c^2 + d^2 = 4.
Prove that there are two of aa, bb, cc, dd with sum greater or equal to 22.

Solution

Without loss of generality, let abcda \ge b \ge c \ge d and we will prove that a+b2a+b \ge 2.
We have that abc2ab \ge c^2 and abd2ab \ge d^2, so by adding them we have: 2abc2+d22ab \ge c^2 + d^2.
Therefore,
(a+b)2=a2+b2+2aba2+b2+c2+d2=4, (a+b)^2 = a^2 + b^2 + 2ab \ge a^2 + b^2 + c^2 + d^2 = 4,
so a+b2a+b \ge 2.

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