<s>[INST] Find maximum of $P$ Let $$P = \frac{{{x^2}}}{{{x^2} + yz + x + 1}} + \frac{{y + z}}{{x + y + z + 1}} - \frac{{1 + yz}}{9}.$$ 
Find maximum of $P$  where $x, y,z$ are nonnegative real numbers such that ${x^2} + {y^2} + {z^2} = 2$.
I guess $\max P=\dfrac{5}{9}$ when $x=y=1, z=0$. But I can not prove.
 [/INST] Step 1. Prove that
$$
x^2+yz+x+1\geq x(x+y+z+1)
$$
by using the expansion of $(x-y-z)^2\geq 0$ and $x^2+y^2+z^2=2$.
Step 2. Observe that
$$
x+y+z\leq\sqrt{2(x^2+(y+z)^2)}=2\sqrt{1+yz}
$$
Step 3. By Step 1, we have
$$
P\leq\frac{x^2}{x(x+y+z+1)}+\frac{y+z}{x+y+z+1}-\frac{1+yz}{9}=1-Q
$$
with 
$$
Q=\frac{1}{x+y+z+1}+\frac{1+yz}{9}.
$$
By step 2 we have
$$
Q\geq \frac{1}{2\sqrt{1+yz}+1}+\frac{1+yz}{9}.
$$
Let $t=\sqrt{1+yz}\geq 1$ and
$$
f(t)=\frac{1}{2t+1}+\frac{t^2}{9}
$$
We can find
$$
\min_{t\geq 0}f(t)=f(0)=\frac{4}{9}
$$
and so  $\max P=\frac{5}{9}$ at $x=y=1, z=0$.
</s>