2015/021) If $x^3 + y^3 + z^3 = 81$ ,then prove that $x + y +z\le 9$

Applying C00auchy schwarz inequality to (x,y,z) and (1,1,1) we have
$(x+y+z)^2 \le (1+1+1)(x^2+y^2+z^2)$
or $(x+y+z)^2 \le 3(x^2+y^2+z^2)$
hence $(x+y+z)^4 \le 9(x^2+y^2+z^2)^2$
applying same inequality to $\sqrt{x},\sqrt{y},\sqrt{z}$ and  $\sqrt{x^3},\sqrt{y^3},\sqrt{z^3}$

 we get :
$(x^2+y^2+z^2)^2\le (x+y+z)(x^3+y^3+z^3)$
or $(x^2+y^2+z^2)^2 \le 81(x+y+z) \cdots (2)$

 From (1) and (2)...
$(x+y+z)^4 \le 9 * 81 (x+y+z)$

or $(x+y+z)^3\le 9^3$

or $(x+y+z)\le 9$
refer to https://in.answers.yahoo.com/question/index?qid=20120510235300AA1XWbg