Докажите неравенства. а) (х+1)²> х(х+2) б) а²+1> =2(3а-4)

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1. найдем производную. y'=6x²-6x-72=6(x²-x-12)

2. найдем стационарные точки. 6(x²-x-12)=0, по Виету х=4; х=-3.

3. Выясним, как ведет себя производная при переходе через эти точки.  решив неравенство, например, y'>0, методом интервалов.

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точка х=-3- точка максимума, максимум равен у(-3)=2*(-3)³ - 3*(-3)²- 72*(-3) + 5=-54-27+216+5=221-81=140

точка х=4- точка минимума, минимум функции равен

у(4)=2*4³ - 3*4²- 72*4 + 5=128-48-228+5=-143

Интервалы монотонности - убывает функция при х∈[-3;4]

(-∞;-3] и при х∈[4;+∞)