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[不等式] 不等式恒成立下的条件的\(b+3a\)最小值

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isee posted 2021-8-20 23:47 |Read mode
已知$a>1$,\(b\in R\),当$x>0$时,$\left[ (a-1)x-1 \right]\cdot \left( \frac {x^2-4}{2x}-b \right)\ge 0$恒成立,则\(b+3a\)的最小值是__`\sqrt 2+3`____.

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