What is the min value

Differenting the given constraint we get $dx+dy+dz=0$

For $f(x,y,z)=\dfrac {x^2}{1+x^2}+\dfrac {y^2}{1+y^2}+\dfrac {z^2}{1+z^2}$ to be maximum,

$\dfrac {2x}{(1+x^2)^2}dx+\dfrac {2y}{(1+y^2)^2}dy+\dfrac {2z}{(1+z^2)^2}dz=0$ .

Applying LaGrange's method of undetermined multipliers and using the constraint $x+y+z=1$ , we get $x=y=z=\dfrac {1}{3}$

The maximum value of the given expression is $3\times \dfrac {\frac{1}{9}}{1+\frac{1}{9}}=\dfrac {3}{10}=\boxed {0.3}$ .

Solution to this problem :

One obvious number for which $\sqrt n+\sqrt {n-64}$ is an integer is $64$ .

For other values, let $n=m^2$ , where $m$ is an integer. Then

$m^2-64$ must be a perfect square, say, $p^2$ , where $p$ is an integer. So $p, 8,m$ form a Pythagorean triplet. The primitive one is $8,15,17$ , while the nonprimitive one is $6,8,10$ . So there are two other values of $n: 17^2=289$ and $n=10^2=100$ .

Therefore in all, there are $\boxed 3$ values of $n$ satisfying the given condition.

We note that $f(x) = \dfrac {x^2}{1+x^2}$ is convex within $x \in [0,1]$ . By Jensen's inequality ,

$\begin{aligned} \frac 13 \left(\frac {x^2}{1+x^2} + \frac {y^2}{1+y^2} + \frac {z^2}{1+z^2} \right) & \ge \frac {\left(\frac {x+y+z}3\right)^2}{1+ \left(\frac {x+y+z}3\right)^2} = \frac {\frac 19}{1+\frac 19} = \frac 1{10} \\ \implies \frac {x^2}{1+x^2} + \frac {y^2}{1+y^2} + \frac {z^2}{1+z^2} & \ge \frac 3{10} = \boxed{0.3} \end{aligned}$