Biến đổi như sau $$\dfrac{bc}{6}+\dfrac{ac}{3}+\dfrac{ab}{2}=1 \leftrightarrow \dfrac{b}{2}.\dfrac{c}{3}+\dfrac{c}{3}.\dfrac{a}{1}+\dfrac{a}{1}.\dfrac{b}{2}=1$$
Đặt $(\dfrac{a}{1},\dfrac{b}{2},\dfrac{c}{3})=(x,y,z), x,y,z>0 \rightarrow xy+yz+zx=1$
Mặt khác $$A=\dfrac{1}{a^2+1}+\dfrac{1}{(\dfrac{b}{2})^2+1}+\dfrac{1}{(\dfrac{c}{3})^2+1}=\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}+\dfrac{1}{z^2+1}$$
Do đó ta cần tìm max của $$\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}+\dfrac{1}{z^2+1}$$
Với $$xy+yz+zx=1$$
Thật vậy thay
$$1=xy+yz+zx \rightarrow A=\sum{\dfrac{1}{x^2+xy+yz+zx}}=\sum{\dfrac{1}{(x+y)(y+z)}}=\dfrac{(x+y)+(y+z)+(z+x)}{(x+y)(y+z)(z+x)}=\dfrac{2(x+y+z)}{(x+y)(y+z)(z+x)}$$
Áp dụng bdt $(x+y)(y+z)(z+x)\geq \dfrac{8}{9}(x+y+z)(xy+yz+xz)$
Suy ra $A\le \dfrac{2(x+y+z)}{\dfrac{8}{9}(x+y+z)(xy+xz+zx)}$ thay $xy+yz+zx=1 \rightarrow A\le \dfrac{9}{4}$
Dấu $= \leftrightarrow x=y=z=\sqrt{\dfrac{1}{3}} \rightarrow a=..., b=...,c=...$ Làm tiếp hộ mình
