expr=(-a^4 c-a^3 b c+a^2 b^2 c+a b^3 c+a^3 c^2+3 a^2 b c^2+a b^2 c^2+a^2 c^3-a b c^3-a c^4)/(a^2 b^3+2 a b^4+b^5+2 a^2 b^2 c+4 a b^3 c+2 b^4 c+a^2 b c^2+2 a b^2 c^2+b^3 c^2)

newExpr=expr/. {a->x+y,b->y+z,c->z+x};


expandedExpr=Expand[newExpr]

Simplify[expandedExpr]

(*定义原始表达式*)expr=((x+y) (x+z) (x^2 (y+z)+9 y z (y+z)+x (y^2+10 y z+z^2)))/((y+z) (x^2+2 y^2+5 y z+2 z^2+3 x (y+z))^2);

(*完全轮换替换规则*)
cyclicReplacement={x->y,y->z,z->x};

(*应用替换规则*)
result=expr/. cyclicReplacement

(*定义原始表达式*)expr=(((x+y) (y+z) ((y^2) (x+z)+9 x z (x+z)+y ((x^2)+10 x z+(z^2))))/((x+z) (((2 (x^2)+(y^2)+5 x z+2 (z^2)+3 y (x+z)))^2)));

(*完全轮换替换规则*)
cyclicReplacement={x->y,y->z,z->x};

(*应用替换规则*)
result=expr/. cyclicReplacement

expr=(2 u v t)/((u+v-t) (u+t-v) (t+v-u))
newExpr=expr/. {u->x+y,v->y+z,t->z+x};

(*展开新表达式*)
expandedExpr=Expand[newExpr]

Simplify[expandedExpr]

(*定义函数*)f[x_]:=1+(1/(2x))+(2/(2+3x))

(* 代入 f(x)*)
result=f[(((x+y) (x+z) (y+z))/(4 x y z))]

(*输出结果*)
result

u=(((x+y) (x+z) ((x^2) (y+z)+9 y z (y+z)+x ((y^2)+10 y z+(z^2))))/((y+z) ((((x^2)+2 (y^2)+5 y z+2 (z^2)+3 x (y+z)))^2)))
v=(((x+y) (y+z) ((y^2) (x+z)+9 x z (x+z)+y ((x^2)+10 x z+(z^2))))/((x+z) (((2 (x^2)+(y^2)+5 x z+2 (z^2)+3 y (x+z)))^2)))
w=(((x+z) (y+z) (9 x y (x+y)+((x^2)+10 x y+(y^2)) z+(x+y) (z^2)))/((x+y) (((2 (x^2)+5 x y+2 (y^2)+3 (x+y) z+(z^2)))^2)))

ClearAll[p]
Subscript[p,1]
Subscript[p,2]

(Subscript[p, 1])=1+((2 x y z)/((x+y) (x+z) (y+z)))+(2/(2+((3 (x+y) (x+z) (y+z))/(4 x y z))))

(*定义函数*)f[x_]:=(3x+6)/(3x+2)

(* 代入 f(x)*)
result=f[(((x+y) (x+z) (y+z))/(4 x y z))]

(Subscript[p, 2])=((6+((3 (x+y) (x+z) (y+z))/(4 x y z)))/(2+((3 (x+y) (x+z) (y+z))/(4 x y z))))

Subscript[s,11]
Subscript[s,12]
Subscript[s,10]

Subscript[s,21]
 Subscript[s,22]
Subscript[s,20]

(Subscript[s, 11])=Together[(Subscript[p, 1])-u-v-w]
(Subscript[s, 12])=Together[(Subscript[s, 11])^2-4(u v+v w+w u)]
(Subscript[s, 10])=Together[(Subscript[s, 12])^2-64 u v w (Subscript[p, 1])]

(*由破根号定理可知，若out[260]>=0则内心三角形不等式下界成立*)

(Subscript[s, 21])=Together[(Subscript[p, 2])-u-v-w]
(Subscript[s, 22])=Together[(Subscript[s, 21])^2-4(u v+v w+w u)]
(Subscript[s, 20])=Together[(Subscript[s, 22])^2-64 u v w (Subscript[p, 2])]

(*由破根号定理可知，若out[260]>=0则内心三角形不等式上界成立*)