Algebra Problem No.2

Note that $x^2 +y^2 +z^2 - 2x+2=(x-1)^2 +y^2 +z^2 +1$ . By Titu's lemma , $(x-1)^2 +y^2 +z^2 +1 \ge \dfrac {(x-1)^2 +y+z+1)^2} {1+1+1+1} =\dfrac{(x+y+z) ^2} 4$ . Equality occurs when $x=2$ and $y=z=1$ . Then $(x-1)^2 +y^2 +z^2 +1 \ge 4$ .

By AM-GM inequality , $x + y +1 +z+1 \ge 3\sqrt[3]{x(y+1)(z+1)}$ . Equality occurs also when $x=2$ and $y=z=1$ . Then $x(y+1)(z+1) \le 8$ .

Therefore $P \le \dfrac 1{\sqrt 4}-\dfrac 28 =\dfrac 14 =\boxed {0.25}$ .

For $P$ to be maximum, $\dfrac {\partial P}{\partial x}=0$

$\implies \dfrac {2}{x(y+1)(z+1)}=\dfrac {x(x-1)}{α^{\frac 32}}$ ,

where $α=x^2+y^2+z^2-2x+2$

$\dfrac {\partial P}{\partial y}=0$

$\implies \dfrac {2}{x(y+1)(z+1)}=\dfrac {y(y+1)}{α^{\frac 32}}$

$\dfrac {\partial P}{\partial z}=0$

$\implies \dfrac {2}{x(y+1)(z+1)}=\dfrac {z(z+1)}{α^{\frac 32}}$

Hence, $x(x-1)=y(y+1)=z(z+1)\implies y=z=x-1$

So $\dfrac {2}{x(y+1)(z+1)}=\dfrac {y(y+1)}{α^{\frac 32}}$

$\implies \dfrac {2}{(y+1)^3}=\dfrac {y(y+1)}{(3y^2+1)^{\frac 32}}$

Solving this equation we get $y=1$

So, $z=1,x=1+1=2$

Therefore $P_{\max}=\dfrac {1}{\sqrt {3\times 1^2+1}}-\dfrac 28=\boxed {0.25}$ .