Why must there be $\sqrt {10}$ ?

Given is an equation of sphere

$x^2+y^2+z^2=1$

We write it in parametric form as

$x=\cos \theta$ $\sin \phi$ , $y=\cos \phi$ , $z=\sin \theta$ $\sin \phi$ , where $\theta$ and $\phi$ are independent of each other.

After plugging these values in given expression, it can be rewritten as

$\sin \phi$ $\cos \phi$ $(\sqrt{10}\cos \theta+\sin \theta)$ ,

which clearly (crystal clear to a Level 5 user) has a maximum value equal to

$\frac {1}{2}$ $\sqrt{10+1}$ $=$ $\frac{\sqrt{11}}{2}$ , which is approximately equal to $\boxed{1.658}$ .

My solution is based on Cauchy-Schwarz Inequality .

$x^2+y^2+z^2=1$

$x^2+z^2=1-y^2$

By Cauchy-Schwarz Inequality,

$(x^2+z^2)(10+1)\geq(\sqrt{10}x+z)^2$

$\sqrt{11(1-y^2)}\geq\sqrt{10}x+z$

Now, the problem can be written as to find the maximum value of

$y\sqrt{11(1-y^2)}$

$\sqrt{11(y^2-y^4)}$

By simple derivatives, the maximum value of $y^2-y^4$ for $y$ is a real number is $\frac{1}{4}$ .

Substitute to the expression and we will get

$\frac{\sqrt11}{2}\approx 1.658$

Firstly we rewrite the ugly expression as an even uglier one: $\sqrt{10}xy+yz=11^{\frac{1}{4}}x\frac{10^{\frac{1}{2}}}{11^{\frac{1}{4}}}y+\frac{1}{11^{\frac{1}{4}}}y11^{\frac{1}{4}}z$

Now we use the fact that $ab\leq\frac{a^{2}+b^{2}}{2}$ for all real $a, b$ .

Letting $a=11^{\frac{1}{4}}x, b=\frac{10^{\frac{1}{2}}}{11^{\frac{1}{4}}}y$ Then $11^{\frac{1}{4}}x\frac{10^{\frac{1}{2}}}{11^{\frac{1}{4}}}y$ is less than or equal to $\frac {11^{\frac{1}{2}}x^{2}+\frac{10}{11^{\frac{1}{2}}}y^{2}}{2}$

Similarly, $\frac{1}{11^{\frac{1}{4}}}y11^{\frac{1}{4}}z\leq\frac {\frac{1}{11^{\frac{1}{2}}}y^{2}+11^{\frac{1}{2}}z^{2}}{2}$

Adding both up, we get $\sqrt{10}xy+yz\leq\frac{\sqrt{11}x^{2}+\sqrt{11}y^{2}+\sqrt{11}z^{2}}{2}=\frac{\sqrt{11}}{2}$ This is approximately 1.658. Also, equality can be achieved when $x=\sqrt {\frac {5}{11}}, y=\sqrt {\frac {1}{2}}, z=\sqrt {\frac {1}{22}}$ Hence answer is 1.658.

Using the method of Lagrange multiplier we get, $\sqrt{10} y = 2xL$ ( Let $L$ denote the multiplier/the symbol lambda). Also, we get $\sqrt10x + z = 2yL$ and $y=2zL$ . From these equations by substituting for $y$ we know that $\sqrt10z=x$ . By some simple substitution you can find the value of $L$ to be $\sqrt{11} /2$ . Then use the fact that $y= 2z\times \sqrt{11}/2$ and $\sqrt10z=x$ to get $x$ in terms of $y$ . This will give $x=\sqrt{10}y/\sqrt{11}$ . Knowing these two last equations we can substitute for $y$ and $z$ in the equation given in the question. Solving this new equation will give $x=\sqrt{10/22}$ . Using this value of $x$ you can find the values of $z$ and $y$ which are $\sqrt{1/22}$ and $\sqrt{11/22}$ respectively. Then just substitute the values into the expression to get a maximum value of $\boxed{1.658}$

Consider $U=\sqrt{11}\left(\sqrt{10}xy+yz\right)=\left(\sqrt{11}y\right)\left(\sqrt{10}x+z\right)$ .

By AM-GM Inequality, $U \le \frac{11y^2+\left(\sqrt{10}x+z\right)^2}{2}$ .

By Cauchy-Schwartz Inequality, $\left(x^2+z^2\right)(10+1) \ge \left(\sqrt{10}x+z\right)^2$ .

Hence, $U\le \frac{11y^2+11\left(x^2+z^2\right)}{2}=\frac{11}{2}$ .

Now, $\sqrt{10}xy+yz =\frac{U}{\sqrt{11}}\le \frac{\sqrt{11}}{2} \approx 1.658$ .

The minimum is attained if $\begin{cases}\frac{x^2}{10} &=&z^2 \\ \sqrt{11}y &=& \sqrt{10}x+z\end{cases}$ , which eventually means that $\begin{cases} x &=& \sqrt{\frac{5}{11}} \\ y &=& \frac{1}{\sqrt{2}} \\ z &=& \frac{1}{\sqrt{22}} \end{cases}$ .