# Determine the real numbers, x,y,z, so that sqrt(x^2-2x+5) + sqrt(y^2-4y+5) + sqrt(z^2+6z+18)=6

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### 1 Answer

We'll try to write the folowing expressions in this way:

x^2-2x+5=x^2-2x+1+4=(x^2-2x+1)+4=(x-1)^2+4

y^2-4y+5=y^2-4y+4+1=(y^2-4y+4)+1=(y-2)^2+1

###### z^2+6z+18=z^2+6z+9+9=(z^2+6z+9)+9=(z+3)^2+9

We've put the expressions above in this manner of writting for highlighting binomial expressions.

sqrt [(x-1)^2+4]>sqrt 4=2

sqrt [(y-2)^2+1]>sqrt 1=1

sqrt[(z+3)^2+9]>sqrt 9=3

If we'll add these expressions, the result will be:

sqrt [(x-1)^2+4]+sqrt [(y-2)^2+1]+sqrt[(z+3)^2+9]>2+1+3

sqrt [(x-1)^2+4]+sqrt [(y-2)^2+1]+sqrt[(z+3)^2+9]>6 **q.e.d**