I'm trying to solve this extremal problem:
\begin{align} &xy + yz \rightarrow \max \\ &x^2 + y^2 = 2 \\ &y + z = 2 \end{align} So, using AM-GM inequality: \begin{align} (\sqrt{xy})^2 + (\sqrt{yz})^2 &\leq \Big(\dfrac{x + y}{2}\Big)^2+\Big(\dfrac{y + z}{2}\Big)^2 \\ &=\Big(\dfrac{x^2 + y^2 + 2xy}{4}\Big)+1 \\ &=\Big(\dfrac{1}{2} + \dfrac{xy}{2}\Big) + 1 \end{align}
So what am i do the next step? or it can be an answer?
On
Eliminate $z$: we're maximizing $y(x+2-y)$ subject to $x^2+y^2=2$. Now use Lagrange multipliers:$$L=xy+2y-y^2+\lambda(2-x^2-y^2)\implies 0=L_x=y-2\lambda x,\,0=L_y=x+2-2y-2\lambda y.$$So$$y=2\lambda x,\,0=2+x(1-4\lambda-4\lambda^2)\implies x=\frac{2}{4\lambda^2+4\lambda-1},\,y=\frac{4\lambda}{4\lambda^2+4\lambda-1}.$$We seek positive $\lambda$ by solving$$2=\frac{4(1+4\lambda^2)}{(4\lambda^2+4\lambda-1)^2}\implies0=(4\lambda^2+4\lambda-1)^2-2(1+4\lambda^2)=16\lambda^4+32\lambda^3-8\lambda-1.$$There's only one such root:$$\lambda=\tfrac12\implies x=y=z=1,\,y(x+z)=2.$$
Alternative approach: write $x=\sqrt{2}\sin t,\,y=\sqrt{2}\cos t,\,t\in[0,\,\pi/2]$ so we want to maximize$$2\cos t(\sin t+\sqrt{2}-\cos t)=2\sqrt{2}\cos t(1+\sin(t-\pi/4)).$$With $u=t-\pi/4\in[-\pi/4,\,\pi/4]$, this becomes$$2(\cos u-\sin u)(1+\sin u)=2-3u^2+o(u^2).$$
I suppose i have smoothest solution to my problem(without using trigonometry etc.): $(x - y)^2 \ge 0 \Rightarrow x^2 + y^2 \ge 2xy \Rightarrow 2 \ge 2xy \Rightarrow xy \le 1$ and general maximum is $2$.