a, [tex]3-2VT=\frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2a}[/tex] [tex]= \frac{a^2}{a^2+2ab}+\frac{b^2}{b^2+2bc}+\frac{c^2}{c^2+2ca} \geq \frac{(a+b+c)^2}{(a+b+c)^2} =1[/tex]
[tex]\Rightarrow VT \leq 1[/tex]
đẳng thức xảy ra <=> a=b=c>0
b,[tex]3-3VT=\sum \frac{2a+c}{2a+3b+c}[/tex]
[tex]\sum \frac{2a}{2a+3b+c}=2\sum \frac{a^2}{2a^2+3ab+ac}\geq 2.\frac{(a+b+c)^2}{2(a+b+c)^2}=1[/tex]
[tex]\sum \frac{c}{2a+3b+c}=\sum \frac{c^2}{2ac+3bc+c^2}\geq \frac{(a+b+c)^2}{(a+b+c)^2+3(ab+bc+ca)}[/tex]
[tex]3(ab+bc+ca)\leq (a+b+c)^2[/tex]
[tex]\Rightarrow \sum \frac{c}{2a+3b+c}\geq \frac{1}{2}[/tex]
[tex]\Rightarrow 3-3VT\geq \frac{3}{2} \Rightarrow VT\leq \frac{1}{2}[/tex]
đẳng thức xảy ra <=> a=b=c>0
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
