Câu hỏi của Cao Thi Thuy Duong

Question

cho a,b,c >0 chứng minh

$\frac{a}{2b+3c}+\frac{b}{2c+3a}+\frac{c}{2a+3b}\ge\frac{3}{5}$

Nguyễn Việt Lâm · Accepted Answer

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

$P\ge\frac{\left(a+b+c\right)^2}{5\left(ab+bc+ca\right)}\ge\frac{3\left(ab+bc+ca\right)}{5\left(ab+bc+ca\right)}=\frac{3}{5}$

Dấu "=" xảy ra khi $a=b=c$Câu trả lời: $P=\frac{a^2}{2ab+3ac}+\frac{b^2}{2bc+3ab}+\frac{c^2}{2ac+3bc}$

$P\ge\frac{\left(a+b+c\right)^2}{5\left(ab+bc+ca\right)}\ge\frac{3\left(ab+bc+ca\right)}{5\left(ab+bc+ca\right)}=\frac{3}{5}$

Dấu "=" xảy ra khi $a=b=c$