Put \(\left\{{}\begin{matrix}x=b+c-a>0\\y=c+a-b>0\\z=a+b-c>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{y+z}{2}\\b=\dfrac{x+z}{2}\\c=\dfrac{x+y}{2}\end{matrix}\right.\)

Instead we have:A=\(\dfrac{y+z}{2x}+\dfrac{z+x}{2y}+\dfrac{x+y}{2z}=\dfrac{1}{2}\left(\dfrac{y}{x}+\dfrac{z}{x}+\dfrac{z}{y}+\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{y}{z}\right)\)

Applying Cauchy's theorem for 2 numbers not negative we have:

A\(=\dfrac{1}{2}\left(\dfrac{y}{x}+\dfrac{x}{y}+\dfrac{z}{x}+\dfrac{x}{z}+\dfrac{y}{z}+\dfrac{z}{y}\right)\)\(\ge\dfrac{1}{2}\left(2\sqrt{\dfrac{y}{x}.\dfrac{x}{y}}+2\sqrt{\dfrac{z}{x}.\dfrac{x}{z}}+2\sqrt{\dfrac{y}{z}.\dfrac{z}{y}}\right)\)

\(=\dfrac{1}{2}\left(2+2+2\right)=\dfrac{1}{2}.6=3\)

\(\Rightarrow\)The problem must prove and A=3 when x=y=z

Put ⎧⎪⎨⎪⎩x=b+c−a>0y=c+a−b>0z=a+b−c>0⇒⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩a=y+z2b=x+z2c=x+y2

Instead we have:A=y+z2x+z+x2y+x+y2z=12(yx+zx+zy+xy+xz+yz)

Applying Cauchy's theorem for 2 numbers not negative we have:

A=12(yx+xy+zx+xz+yz+zy)

≥12(2√yx.xy+2√zx.xz+2√yz.zy)

=12(2+2+2)=12.6=3

⇒

The problem must prove and A=3 when x=y=z

Put ⎧⎪⎨⎪⎩x=b+c−a>0y=c+a−b>0z=a+b−c>0⇒⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩a=y+z2b=x+z2c=x+y2

Instead we have:A=y+z2x+z+x2y+x+y2z=12(yx+zx+zy+xy+xz+yz)

Applying Cauchy's theorem for 2 numbers not negative we have:

A=12(yx+xy+zx+xz+yz+zy)

≥12(2√yx.xy+2√zx.xz+2√yz.zy)

=12(2+2+2)=12.6=3

⇒

The problem must prove and A=3 when x=y=z