blue2000

7. Chứng minh rằng với mọi số thực dương $a,b,c$ ta có:

b) $\frac{1}{a^{3}+8abc}+\frac{1}{b^{3}+8abc}+\frac{1}{c^{3}+8abc}\leq \frac{1}{3abc}$

BPT đã cho

$\Leftrightarrow \frac{bc}{a^{2}+8bc}+\frac{ca}{b^{2}+8ca}+\frac{ab}{c^{2}+8ab}\leq \frac{1}{3}$

$\Leftrightarrow (\frac{1}{8}-\frac{bc}{a^{2}+8bc})+(\frac{1}{8}-\frac{ca}{b^{2}+8ca})+(\frac{1}{8}-\frac{ab}{c^{2}+8ab})\geq \frac{1}{24}$

$\Leftrightarrow \frac{a^{2}}{a^{2}+8bc}+\frac{b^{2}}{b^{2}+8ca}+\frac{c^{2}}{c^{2}+8ab}\geq \frac{1}{3}$

Mặt khác

$\frac{a^{2}}{a^{2}+8bc}+\frac{b^{2}}{b^{2}+8ca}+\frac{c^{2}}{c^{2}+8ab}$

$\geq \frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}+8(ab+bc+ca)}$

$=\frac{(a+b+c)^{2}}{(a+b+c)^{2}+6(ab+bc+ca)}$

$\geq \frac{(a+b+c)^{2}}{(a+b+c)^{2}+2(a+b+c)^{2}}=\frac{1}{3}$ (đpcm)

3. Chứng minh rằng với mọi số thực không âm $a,b,c$ ta có:

$a\sqrt{a^{2}+8bc}+b\sqrt{b^{2}+8ca}+c\sqrt{c^{2}+8ab}\leq (a+b+c)^{2}$

$a\sqrt{a^{2}+8bc}+b\sqrt{b^{2}+8ca}+c\sqrt{c^{2}+8ab}$

$=\sqrt{a}\sqrt{a^{3}+8abc}+\sqrt{b}\sqrt{b^{3}+8abc}+\sqrt{c}\sqrt{c^{3}+8abc}$

$\leq \sqrt{(a+b+c)(a^{3}+b^{3}+c^{3}+24abc)}$

Cần chứng minh:

$a^{3}+b^{3}+c^{3}+24abc\leq (a+b+c)^{3}$

$\Leftrightarrow 24abc\leq3(a+b)(b+c)(c+a)$

$\Leftrightarrow 8abc\leq(a+b)(b+c)(c+a)$ (đpcm)