\(\text{Condition}:1-2x-2x^2\ge0\)
We have:
\(A=x+\sqrt{1-2x-2x^2}\)
\(\Rightarrow M=-2A=-2x-2\sqrt{1-2x-2x^2}\)
Now we need to find min of M
We have it:
\(M=-2x-2\sqrt{1-2x-2x^2}=\left(1-2x-2x^2-2\sqrt{1-2x-2x^2}+1\right)+2x^2-2=\left(\sqrt{1-2x-2x^2}-1\right)^2+2x^2-2\ge-2\)
\(\Rightarrow-2A\ge-2\Leftrightarrow A\le1\)
Sign '=' happening when \(x=0\)