# Prove the induction: (1+x1)(1+x2)...(1+xn)>=1+x1+x2+..+xn,xk>-1Solution(1+x1)(1+x2)...(1+xn)*(1+ xn+1)>=(1+x1+x2+....+xn)*(1+...

Prove the induction: (1+x1)(1+x2)...(1+xn)>=1+x1+x2+..+xn,xk>-1

Solution

(1+x1)(1+x2)...(1+xn)*(1+ xn+1)>=(1+x1+x2+....+xn)*(1+ xn+1)=

(1+x1+x2+....+xn)+xn+1+(x1+x2+...+xn)*xn+1>=1+x1+x2....xn+1

I don't understand why this (1+x1+x2+....+xn)+xn+1, if we multiply, then we should get just (1+x1+x2+....+xn)...

### 2 Answers | Add Yours

You should come up with this substitution:

`1+x_1+x_2+....+x_n = y`

Multiplying y by `(1 + x_(n+1))` yields:

`y* (1 + x_(n+1)) = y + y*x_( n+1)`

Notice that `x_1+x_2+....+x_n = y-1`

Notice that multiplying `x_( n+1)` by each term of y yields:

`y* (1 + x_(n+1)) = y +x_(n+1) + (y-1)*x_(n+1) + ...+(y-1)*x_(n+1)`

Plugging `1+x_1+x_2+....+x_n` in the equation above and factorizing by `x_(n+1)` yields:

`1+x_1+x_2+....+x_n +x_(n+1) + (x_(n+1))(x_1+x_2+....+x_n)`

**The expression `(1+x_1+x_2+....+x_n)*(1+ x_(n+1))=1+x_1+x_2+....+x_n +x_(n+1) + (x_(n+1))(x_1+x_2+....+x_n)` is verified.**

Thanks, aldo it's a little complicated, it helped! ;)