$x,y\in\Bbb R,$ find the maxima of $\frac{x+2y+3}{\sqrt{x^2+y^2+1}}$

Question

I want to use Cauchy–Schwarz inequality, I sqared $\frac{x+2y+3}{\sqrt{x^2+y^2+1}}$ and got $\frac{x^2+4y^2+9}{x^2+y^2+1}$, not sure if I am doing fine

Answer 1

Not that $$(x+2y+3)^2\le (x^2+y^2+1)(1+4+9)$$

Answer 2

By C-S $$\frac{x+2y+3}{\sqrt{x^2+y^2+1}}\leq\sqrt{\frac{14(x+2y+3)^2}{(1^2+2^2+3^2)(x^2+y^2+1)}}\leq$$ $$\leq\sqrt{\frac{14(x+2y+3)^2}{(x+2y+3)^2}}=\sqrt{14}.$$ The equality occurs for $(x,y,1)||(1,2,3)$ and $x+2y+3>0,$ id est, for $(x,y)=\left(\frac{1}{3},\frac{2}{3}\right),$

which says that we got a maximal value.