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攒拳怒目的坚果
9月前
严格的证明是：
$由g(a)的定义知g(a)≥f(1)=|1-a|,g(a)≥f(\sqrt{2}-1)=|3-2\sqrt{2}-(\sqrt{2}-1)a|$
$故\sqrt{2}g(a)≥(\sqrt{2}-1) \cdot |1-a|+1 \cdot |3-2\sqrt{2}-(\sqrt{2}-1)a|$
$=|(\sqrt{2}-1)-(\sqrt{2}-1)a|+|3-2\sqrt{2}-(\sqrt{2}-1)a|≥|((\sqrt{2}-1)-(\sqrt{2}-1)a)-(3-2\sqrt{2}-(\sqrt{2}-1)a)|=3\sqrt{2}-4$
$故g(a)≥\frac{3\sqrt{2}-4}{\sqrt{2}}=3-2\sqrt{2}$
$而g(2\sqrt{2}-2)=3-2\sqrt{2},故g(a)最小值为3-2\sqrt{2},取等时第二行两个不等式均取等,必有a=2\sqrt{2}-2$