shmm wrote: Find the max and minimum without using dervivate: $\sqrt{x} + \sqrt{x-\frac{1}{2}}$ This expression is defined only for $x\in[\frac 12,+\infty)$ and is increasing. So min is reached when $x=\frac 12$ and is $\frac{\sqrt 2}2$ And max is reached when $x\to+\infty$ and is $+\infty$

sorry , edited.

shmm wrote: Find the max and minimum without using dervivate: $\sqrt{x} +4 \cdot \sqrt{\frac{1}{2} - x}$ Let $u=\arcsin \frac 1{\sqrt{17}}$ ,and $x=\frac{\cos^2t}2$ Expression is $\frac{\sqrt{17}}{\sqrt 2}\sin(t+u)$ Maximum is reached when $u+t=\frac{\pi}2$ and is $\boxed{\text{max }=\frac{\sqrt{17}}{\sqrt 2}}$ Minimum is reached either when $t=0$ (and so would be $\frac 1{\sqrt 2}$), either when $t=\frac{\pi}2$ (and so would be $\frac 4{\sqrt 2}$) And so $\boxed{\text{min }=\frac{1}{\sqrt 2}}$

pco's solution by trigonometric function is very nice. My solution Let $f(x)=\sqrt{x}+4\sqrt{\frac{1}{2}-x}\ge 0\left(0\le x\le \frac{1}{2}\right)$. About min $f(x)^2=-15x+8+8\sqrt{x\left(\frac{1}{2}-x\right)}$ $-15x+8\ge \frac{1}{2}$ and $\sqrt{x\left(\frac{1}{2}-x\right)}\ge 0.$Thus $f(x)^2\ge \frac{1}{2}\implies f(x)\ge \frac{1}{\sqrt{2}}\left(=f\left(\frac{1}{2}\right)\right).$ Therefore the min is $\boxed{\frac{1}{\sqrt{2}}}\blacksquare$ About max $f(x)\le \frac{\sqrt{34}}{2}\Leftrightarrow \left(17x-\frac{1}{2}\right)^2\ge 0$.Thus $f(x)\le \frac{\sqrt{34}}{2}\left(=f\left(\frac{1}{34}\right)\right)$ Therefore the max is $\boxed{\frac{\sqrt{34}}{2}}\blacksquare$

we set $f(x)=\sqrt{x}+4 \cdot \sqrt{\frac{1}{2}-x}$ from $C-S$ inequality we have $(f(x))^2 \leqslant \left((\sqrt{x})^{2}+\left(\sqrt{\frac{ 1}{2}-x}\right)^{2}\right)(1^2+4^2) =\left(\frac{1}{2}\right)\cdot (17) \implies f(x) \leqslant \sqrt{\frac{17}{2}}$, so $\max{(f(x))}=\boxed{\sqrt{\frac{17}{2}}}$ Now we also have $0 \leqslant x \leqslant \frac{1}{2}$ so $f(x)^2=8-15x+\sqrt{x \left(\frac{1}{2}-x \right)}$ , now using the bound that $x \leqslant \frac{1}{2}$ we have $\min{(f(x))}=\boxed{\sqrt{\frac{1}{2}}}$