Có $1-x-2x^{2}=(1-2x)(x+1)$
Mặt khác: $1-x-2x^{2}\geq 0\rightarrow -1\leq x\leq\frac{1}{2}\rightarrow 1-2x\geq 0;x+1\geq 0$
Áp dụng BDDT AM-GM ta có:$\sqrt(1-x-2x^{2})\leq \frac{(1-2x)+(x+1)}{2}= \frac{2-x}{2}$
Vậy f(x) $\leq \frac{2-x}{2}+\frac{x}{2}=1$
$\rightarrow$ Max f(x)=1$\Leftrightarrow$ 1-2x=x+1$\Leftrightarrow$3x=0$\Leftrightarrow$x=0