Bất đẳng thức...

Question

Cho các số thực dương a,b,c.Chứng minh rằng:

$\sqrt{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}\geq 1+\sqrt[3]{5+ \sqrt{(\Sigma a^{3})(\Sigma \frac{1}{a^{3}}})}$

Bloody's Rose · Answer 1 · 6/17/2016 1:36:24 PM

Ta có:

$\sqrt{(a+b+c)^{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{3}}$

=

$\sqrt{\left[ {\Sigma a^{3}+3(a+b)(b+c)(c+a)} \right].\left[ {\Sigma \frac{1}{a^{3}}+\frac{3(a+b)(b+c)(c+a)}{(abc)^{2}}} \right]}$

$\geq \sqrt{(\Sigma a^{3})(\Sigma \frac{1}{a^{3}}}+\frac{3(a+b)(b+c)(c+a)}{abc}$

Mà

$\frac{(a+b)(b+c)(c+a)}{abc}=\frac{(a+b+c)(ab+bc+ca)}{abc}-1=(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1$

Do đó:

$3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-3\geq 3(\Sigma a)(\Sigma \frac{1}{a})-3\sqrt{(\Sigma a)(\Sigma \frac{1}{a})}+6$

Lại có:

$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq9$

$\Rightarrow (\sqrt{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}-1)^{3}\geq 5+\sqrt{(\Sigma a^{3})(\Sigma \frac{1}{a^{3}})}$

$\Rightarrow đpcm$

Dấu''='' xra

$\Leftrightarrow a=b=c$

1 Đáp án