a,b,c là độ dài 3 cạnh tam giác=>a,b,c>0
đặt \(\left\{{}\begin{matrix}a-b+c=x\\b-c+a=y\\c-a+b=z\end{matrix}\right.=>\left\{{}\begin{matrix}x+y=2a\\y+z=2b\\x+z=2c\end{matrix}\right.=>\left\{{}\begin{matrix}a=\dfrac{x+y}{2}\\b=\dfrac{y+z}{2}\\c=\dfrac{x+z}{2}\end{matrix}\right.\)
với \(3a-b+c=2a+a-b+c=x+y+x=2x+y\)
\(3b-c+a=2b+b-c+a=y+z+y=2y+z\)
\(3c-a+b=2c+c-a+b=x+z+z=2z+x\)
\(\dfrac{\dfrac{x+y}{2}}{2x+y}+\dfrac{\dfrac{y+z}{2}}{2y+z}+\dfrac{\dfrac{x+z}{2}}{2z+x}=\dfrac{x+y}{2\left(2x+y\right)}+\dfrac{y+z}{2\left(2y+z\right)}+\dfrac{z+x}{2\left(2z+x\right)}\)
\(=>\dfrac{2x+2y}{2x+y}+\dfrac{2y+2z}{2y+z}+\dfrac{2x+2z}{2z+x}\)\(\)
\(=\dfrac{2x+y}{2x+y}+\dfrac{y}{2x+y}+\dfrac{2y+z}{2y+z}+\dfrac{z}{2y+z}+\dfrac{2z+x}{2z+x}+\dfrac{x}{2z+x}\)
\(=3+\dfrac{x}{2z+x}+\dfrac{y}{2x+y}+\dfrac{z}{2y+z}\)=\(3+\dfrac{x^2}{x^2+2zx}+\dfrac{y^2}{y^2+2xy}+\dfrac{z^2}{z^2+2yz}\left(1\right)\)
áp dụng hệ quả Bunhiacopski
=>(1)\(\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+2xy+2yz+2xz}+3\)
mà \(x^2+y^2+z^2+2xy+2yz+2xz=\left(x+y+z\right)^2\)
=>(1)\(\ge3+1=4\).
=>\(\dfrac{x^2}{x^2+2xz}+\dfrac{y^2}{y^2+2xy}+\dfrac{z^2}{z^2+2yz}\ge4-3=1\)
dấu;=; xảy ra<=>x=y=z<=>a=b=c(vì x+y+z=a+b+c)
BDT trên đc CMInh
