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Algebra Difficulty 5.5 AIME, harder Prove it

4.2 4 ** Given positive numbers a,b,ca, b, c satisfying a+b+c=1a+b+c=1, prove: 4a+1+4b+1+\sqrt{4 a+1}+\sqrt{4 b+1}+ 4c+1>2+5\sqrt{4 c+1}>2+\sqrt{5}

Solution

Parse: Because 04a5+1=(5−0\frac{4 a}{\sqrt{5}+1}=(\sqrt{5}- 1) aa, that is, 4a+1>1+(5−1)a\sqrt{4 a+1}>1+(\sqrt{5}-1) a. Similarly, 4b+1>1+(5−1)b,4c+1>1+\sqrt{4 b+1}>1+(\sqrt{5}-1) b, \sqrt{4 c+1}>1+ (5−1)c(\sqrt{5}-1) c, adding the three inequalities proves the statement.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.