PART 2

Question

Cho 3 số thực dương a,b,c thỏa mãn

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq 16.(a+b+c)$

CM:

$\frac{1}{(a+b+2.\sqrt{a+c})^3}+\frac{1}{(b+c+2.\sqrt{b+a})^3}+\frac{1}{(c+a+2.\sqrt{c+b})^3}\leq \frac{8}{9}$

Nguyễn Nhung · Answer 1 · 7/14/2016 10:19:05 AM

ta có

$(a+b+2\sqrt{a+c})^3>(a+b+\sqrt{2(a+c)}) ^{3}= (a+b+ \sqrt{\frac{a+c}{2}}+\sqrt{\frac{a+c}{2}})^{3} \geq \frac{27}{2}(a+b)(a+c)$

TT suy ra ta cần CM

$\sum \frac{1}{(a+b)(a+c)}\leq12$

$\Leftrightarrow \frac{2(a+b+c)}{(a+b)(b+c)(c+a)}\geq12$

$\Leftrightarrow 6(a+b)(b+c)(c+a)\geq a+b+c$

mà

$9(a+b)(b+c)(c+a)\geq8(a+b+c)(ab+bc+ca)$

như vậy ta cân CM

$ab+bc+ca \geq \frac{3}{16} \Leftrightarrow 16(ab+bc+ca) \geq3$

có

$ab+bc+ca \leq16abc(a+b+c) \leq \frac{16}{3}(ab+bc+ca)^{2} \Rightarrow ab+bc+ca\geq \frac{3}{16}$

Không xảy ra dấu bằng

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