Cách 2:Ta có biến đổi:$\frac{(2a+b+c)^2}{2a^2+(b+c)^2}=1+\frac{4a(b+c)}{2a^2+(b+c)^2}+\frac{2a^2}{2a^2+(b+c)^2}$$\Rightarrow VT=3+\Sigma \frac{4a}{2a^2+(b+c)^2}+\Sigma \frac{2a^2}{2a^2+(b+c)^2}$Ta c/m;$A=\Sigma \frac{4a(b+c)}{2a^2+(b+c)^2}\leq 4$A/d $AM-GM$, ta có: $a^2+\frac{(b+c)^2}{4}\geq a(b+c),$ từ đó:$\frac{4a(b+c)}{(b+c)^2+2a^2}$$\leq \frac{4a(b+c)}{\frac{(b+c)^2}{2}+2a(b+c)}=\frac{8a}{4a+b+c}$Theo $AM-GM,$ có:$2(\frac{a}{a+b+c}+\frac{a}{3a})\geq \frac{8a}{4a+b+c}$$\Rightarrow A\leq \Sigma \frac{8a}{4a+b+c}\leq 2+2=4$ (đpcm)Đẳng thức khi $a=b=c=1.$Tiếp tục c/m:$B=\Sigma \frac{2a^2}{2a^2+(b+c)^2}\leq 1$
Cách 2:Ta có biến đổi:$\frac{(2a+b+c)^2}{2a^2+(b+c)^2}=1+\frac{4a(b+c)}{2a^2+(b+c)^2}+\frac{2a^2}{2a^2+(b+c)^2}$$\Rightarrow VT=3+\Sigma \frac{4a}{2a^2+(b+c)^2}+\Sigma \frac{2a^2}{2a^2+(b+c)^2}$Ta c/m;$A=\Sigma \frac{4a(b+c)}{2a^2+(b+c)^2}\leq 4$A/d $AM-GM$, ta có: $a^2+\frac{(b+c)^2}{4}\geq a(b+c),$ từ đó:$\frac{4a(b+c)}{(b+c)^2+2a^2}$$\leq \frac{4a(b+c)}{\frac{(b+c)^2}{2}+2a(b+c)}$$=2-\frac{(b+c)^2}{\frac{(b+c)^2}{2}+2a(b+c)}$$=2-$$\frac{(b+c)^2}{(b+c)^2+4a(b+c)}$bài toán đưa về c/m:$\Sigma \frac{(b+c)^2}{}$
Cách 2:Ta có biến đổi:$\frac{(2a+b+c)^2}{2a^2+(b+c)^2}=1+\frac{4a(b+c)}{2a^2+(b+c)^2}+\frac{2a^2}{2a^2+(b+c)^2}$$\Rightarrow VT=3+\Sigma \frac{4a}{2a^2+(b+c)^2}+\Sigma \frac{2a^2}{2a^2+(b+c)^2}$Ta c/m;$A=\Sigma \frac{4a(b+c)}{2a^2+(b+c)^2}\leq 4$A/d $AM-GM$, ta có: $a^2+\frac{(b+c)^2}{4}\geq a(b+c),$ từ đó:$\frac{4a(b+c)}{(b+c)^2+2a^2}$$\leq \frac{4a(b+c)}{\frac{(b+c)^2}{2}+2a(b+c)}=\frac{
8a}{4a+b+c
}$Theo $AM-GM,$ có:$2
(\frac{
a}{a+b+c
}+\frac{a}{
3a}
)\geq \fra
c{8a}{4a+b+c}$$
\Rightarrow A\leq \Sigma \frac{
8a}{
4a+b+c
}\leq 2+
2=4
$ (
đpc
m)
Đẳng t
hức khi $a
=b=c=1.$Tiếp tục c/m:$
B=\Sigma \frac{
2a^2}{2a^2+(b+c)^2}
\leq 1$