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Question

cho

$a;b;c>0$ .CMR:

$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geqslant 3\left[ {1+\sqrt[3]{\frac{3(a+b+c)(a+b)(b+c)(a+c)}{(ab+bc+ca)^{2}}}} \right]$

tran85295 · Answer 1 · 7/11/2016 2:05:04 AM

Ta có

$\frac{3(a+b+c)}{(ab+bc+ca)^2} \le \frac 1{abc}$

Và

$(a+b)(b+c)(c+a) \ge 8abc$

$\Longrightarrow VP \le 3+3\sqrt{\frac{(a+b)(b+c)(c+a)}{abc}} \le 3+\frac 34.\frac{(a+b)(b+c)(c+a)}{abc}$

Lại có

$VT=3+\sum_{sym}\frac ab$

Nên chỉ cần cm

$\sum_{sym}\frac ab \ge \frac 34.\frac{(a+b)(b+c)(c+a)}{abc}$

$\Leftrightarrow \sum_{sym}\frac ab-6 \ge \frac{3\Big[(a+b)(b+c)(c+a)-8abc \Big]}{4abc}$

$\Leftrightarrow \sum_{cyc} \frac{(a-b)^2}{3ab} \ge \frac{a(b-c)^2+b(c-a)^2+c(a-b)^2}{4abc}=\sum_{cyc} \frac{(a-b)^2}{4ab}$

$\Leftrightarrow \sum_{cyc}\frac{(a-b)^2}{12ab} \ge0$ (luôn đúng)

$\Rightarrow dpcm...$

Trả lời 11-07-16 02:05 AM

tran85295
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ghi thiếu số 3 ở căn r :)) – ๖ۣۜPXM๖ۣۜMinh4212♓ 11-07-16 07:11 AM

score 10 · Answer 2

$BĐT\Leftrightarrow \frac ab+\frac ba+\frac bc+\frac cb+\frac ac+\frac ca\ge3\sqrt[3]\frac{3abc(a+b+c)(a+b)(b+c)(c+a)}{abc(ab+bc+ca)^2}$

Áp dụng BĐT

$3abc(a+b+c)\le (ab+bc+ca)^2\Rightarrow VP\le3\sqrt[3]\frac{(a+b)(b+c)(c+a)}{abc}$

Có

$(a+b)(b+c)(c+a)\ge8abc$

$\Rrightarrow VP\le\sum\frac32\sqrt[3]\frac{(a+b)^2(b+c)^2(c+a)^2}{a^2b^2c^2}\le\sum\frac{(a+b)^2}{2ab}\le\sum\frac{2(a^2+b^2)}{2ab} =VT$

2 Đáp án