Suppose $k$ be a real number so that the inequality
$\sqrt{y-5}+\sqrt{8-y} \ge k$ should have answer. what is the maximum
possible value of $k$?$1)\sqrt{6}-1\quad\quad2)\frac12+\sqrt{3}\quad\quad3)\sqrt6\quad\quad4)2\sqrt2\quad\quad5)\sqrt6+1$
So I translated the question into finding the minimum value of $f(y)$ where $f(y)=\sqrt{y-5}+\sqrt{8-y}\quad$.(because for $k=\min(f(y))$ the inequality certainly has answer.) therefore we should solve $f'(y)=0$:
$$f'(y)=\frac{1}{\sqrt{y-5}}-\frac{1}{\sqrt{8-y}}=0$$
$$y-5=8-y\rightarrow\quad y=6.5$$
And $f(6.5)=2\sqrt{\frac32}=\sqrt{6}$. so the answer is third choice.
Is my answer right? I know to find maximum or minimum of a function we take derivative of it and equate it to $0$. so here I don't know why $f(6.5)$ is minimum. can you please explain it to me?
EDIT:
I just noticed that $\sqrt6$ is not the minimum because $f(5)=f(8)=\sqrt3<\sqrt6$ maybe it is maximum! so I conclude the answer to this problem should be $\sqrt3$ but I don't have this option in the choices
Best Answer
There is already an answer that clears your question
We can find the best minimum without using derivatives $${(\sqrt{y-5}+\sqrt{8-y})}^2=3+2\sqrt{8-y}\sqrt{y-5}\ge 3$$ because $\sqrt{8-y}\sqrt{5-y}\ge 0$ $$\implies \sqrt{y-5}+\sqrt{8-y}\ge \sqrt{3}$$