# pls,help!! show the inequality f(x,y,z)> or=3 is truef(x,y,z)=squareroot(x^2-2xsinz-4cosz+5)+squareroot(y^2-2ysinz-6cosz+10) i don't have a clue what is "domain of f(x,y,z) is R^3"

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Use the formula which expresses fundamental identity of trigonometry: (sin z)^2+(cos z)^2=1 and write x^2-2xsinz-4cosz+5=x^2-2xsinz-4cosz+4+1=x^2-2xsinz-4cosz+4+(sin z)^2+(cos x)^2

Use perfect squares to express (x^2-2xsinz+(sin z)^2)+((cos z)^2-4cosz+4)=(x-sin z)^2+(cos z -2)^2.

Write y^2-2ysinz-6cosz+10=y^2-2ysinz-6cosz+9+1=y^2-2ysinz-6cosz+9+(sin z)^2+(cos z)^2

Use perfect squares to express (y^2-2ysinz+(sin z)^2)+((cos z)^2-6cosz+9)=(y-sin z)^2+(cos z -3)^2.

Use the results to express f(x,y,z).

f(x,y,z)=sqrt((x-sin z)^2+(cos z -2)^2)+sqrt((y-sin z)^2+(cos z -3)^2).

Use Minkowsky's inequality:

sqrt(m^2+n^2)+sqrt(p^2+q^2)>=sqrt((m^2+p^2)+(n^2+q^2))

Write f(x,y,z) with regard to Minkowsky's inequality:

f(x,y,z)=sqrt((x-sin z)^2+(cos z -2)^2)+sqrt((y-sin z)^2+(cos z -3)^2)