Câu hỏi của Cathy Trang

Question

Cho a,b,c>0 thỏa mãn: a+b+c=ab+bc+ca. Tìm GTNN của P:

$P=\frac{a^2}{a^2+3bc}+\frac{b^2}{b^2+3ca}+\frac{c^2}{c^2+3ab}+\sqrt{a^2+b^2+c^2}$

Nguyễn Việt Lâm · Accepted Answer

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

$\Rightarrow a+b+c\ge3$

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

$P\ge\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+a+b+c}+\frac{1}{\sqrt{3}}\left(a+b+c\right)$

$P\ge1-\frac{1}{a+b+c+1}+\frac{1}{\sqrt{3}}\left(a+b+c\right)\ge1-\frac{1}{3+1}+\frac{1}{\sqrt{3}}.3=\frac{3+4\sqrt{3}}{4}$

Dấu "=" xảy ra khi $a=b=c=1$Câu trả lời: $a+b+c=ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2$

$\Rightarrow a+b+c\ge3$

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

$P\ge\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+a+b+c}+\frac{1}{\sqrt{3}}\left(a+b+c\right)$

$P\ge1-\frac{1}{a+b+c+1}+\frac{1}{\sqrt{3}}\left(a+b+c\right)\ge1-\frac{1}{3+1}+\frac{1}{\sqrt{3}}.3=\frac{3+4\sqrt{3}}{4}$

Dấu "=" xảy ra khi $a=b=c=1$