ĐKBT => $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=1-2x$, $x=ab+bc+ca$
Ta có: $3(ab+bc+ca)<=(a+b+c)^2$ => $x<=\frac{1}{3}$
Mặt khác ta có: $\frac{1}{abc}=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}>=\frac{9}{ab+bc+ca}=\frac{9}{x}$
BĐT cần chứng minh suy ra:
$VT>=\frac{1}{1-2x}+\frac{1}{x}+\frac{1}{x}+\frac{7}{x}>=9+21$
V)
$ab+bc+ac\geq 3\sqrt[3]{a^{2}b^{2}c^{2}}=>\sqrt{(ab+bc+ac)^{3}}\geq \sqrt{27}.abc=>\frac{1}{abc}\geq \sqrt{\frac{27}{(ab+bc+ac)^{3}}}=\frac{3}{ab+bc+ac}.\sqrt{\frac{3}{ab+bc+ac}}$
$ab+bc+ac\leq \frac{(a+b+c)^{2}}{3}=\frac{1}{3}$
$=>P=\frac{1}{a^{2}+b^{2}+c^{2}}+\frac{1}{abc}\geq (\frac{1}{a^{2}+b^{2}+c^{2}}+\frac{2}{ab+bc+ac})+\frac{3}{ab+bc+ac}\sqrt{\frac{3}{ab+bc+ac}}-\frac{2}{ab+bc+ac}\geq \frac{9}{(a+b+c)^{2}}+\frac{2}{ab+bc+ac}(\frac{3}{2}.\sqrt{\frac{3}{ab+bc+ac}}-1)....$
Bài viết đã được chỉnh sửa nội dung bởi nntien: 08-06-2018 - 21:19