Đáp án:
Giải thích các bước giải:
Áp dụng BĐT $AM - GM$ ta có:
$\dfrac{a^5}{(a + b)^{4}} + \dfrac{a + b}{8} = \dfrac{a^5}{(a + b)^{4}} + 4.\dfrac{a + b}{32} $
$ ≥ 5\sqrt[5]{\dfrac{a^5}{(a + b)^{4}}(\dfrac{a + b}{32})^{4}} = 5\sqrt[5]{\dfrac{a^5}{16^{5}}} =\dfrac{5a}{16} (1)$
Dấu $'=' ⇔ \dfrac{a^5}{(a + b)^{4}} = \dfrac{a + b}{32} ⇔ a = b$
Tương tự :
$\dfrac{b^5}{(b + c)^{4}} + \dfrac{b + c}{8} ≥ \dfrac{5b}{16} (2)$.Dấu $'=' ⇔ b = c$
$\dfrac{c^5}{(c + a)^{4}} + \dfrac{c + a}{8} ≥ \dfrac{5c}{16} (3)$.Dấu $'=' ⇔ c = a$
$(1) + (2) + (3) :$
$ \dfrac{a^5}{(a + b)^{4}} + \dfrac{b^5}{(b + c)^{4}} + \dfrac{c^5}{(c + a)^{4}} + \dfrac{a + b + c}{4} ≥ \dfrac{5(a + b + c)}{16}$
$ ⇔ \dfrac{a^5}{(a + b)^{4}} + \dfrac{b^5}{(b + c)^{4}} + \dfrac{c^5}{(c + a)^{4}} ≥ \dfrac{1}{16}(a + b + c) (đpcm)$
Dấu $'=' ⇔ a = b = c$
