Maximizing Radical Sum

The Cauchy-Schwartz Inequality for $2$ terms is, $\left(a^2+b^2\right)\left(c^2+d^2\right)\ge \left(ab+cd\right)^2$ Using the above inequality by setting $(a,b)=\left(\;\sqrt{3x-17},\sqrt{213-3x}\;\right)$ and $(c,d)=\left( 1,\dfrac{1}{\sqrt{3}}\right)$ , we have, $\begin{aligned} \left[\left(\sqrt{3x-17}\right)^2+\left(\sqrt{213-3x}\right)^2 \right]\left[1^2+\left(\frac{1}{\sqrt{3}}\right)^2\right]&\ge \left(\sqrt{3x-17}\cdot 1+\sqrt{213-3x}\cdot \frac{1}{\sqrt{3}} \right)^2 \\ \left[3x-17+213-3x \right]\left[1+\frac{1}{{3}}\right] &\ge \left(\sqrt{3x-17}+\sqrt{71-x}\right)^2 \\ \frac{784}{3}&\ge \left(\sqrt{3x-17}+\sqrt{71-x}\right)^2 \\ \therefore \; \left \lfloor \sqrt{3x-17}+\sqrt{71-x}\right \rfloor &\le \left \lfloor \sqrt{\frac{784}{3}} \right \rfloor =16 \end{aligned}$ The maximum value of the expression $\left \lfloor \sqrt{3x-17}+\sqrt{71-x} \right\rfloor$ is $\boxed{16}$

To further increase the variety of solutions, may I offer the least elegant solution (so far):

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from sympy import *
from sympy.abc import x

def f(x):
    return sqrt(3*x - 17) + sqrt(71 - x)

dif = diff(f(x), x)
xmax = solve(dif, x)[0]
print(floor(f(xmax)))

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16

For the sake of variety I'll present a solution with calculus (which I admit is far less elegant than Sathvik's one). Consider the function $f(x)=\sqrt{3x-17}+\sqrt{71-x}$ Using differentiation rules we have: $f'(x)=\frac{3}{2\sqrt{3x-17}}+\frac{-1}{2\sqrt{71-x}}$ The function's extrema are located in the points, where the derivative is $0$ . Setting the expression for $f'(x)$ equal to zero at $x_0$ gives the equation: $\frac{3}{2\sqrt{3x_0-17}}=\frac{1}{2\sqrt{71-x_0}}\Leftrightarrow 3\sqrt{71-x_0}=\sqrt{3x_0-17}$ After squaring and simplifying, we get $x_0=\frac{164}{3}$ . To determine whether this is maximum or minimum we need to find for which values of $x\ f(x)$ is increasing and for which it is decreasing. This is equivalent to solving the inequalities $f'(x)>0$ and $f'(x)<0$ respectively. After considering the domain of the function the first one yields $\text{Increasing: }x\in\left[\frac{17}{3}; \frac{164}{3}\right)$ and the second one gives $\text{Decreasing: }x\in\left(\frac{164}{3}; 71\right]$ So at $x_0$ the function indeed has its maximum value. It is: $f(x_0)=\sqrt{\frac{784}{3}}\approx16.17$ Our final answer is the floor of the maximum value: $\fbox{16}$ .

$f(x) = \sqrt{3x-17} +\sqrt{71-x}$ $f'(x) = \dfrac{3}{2\sqrt{3x-7}} -\dfrac{1}{2\sqrt{71-x}}$

$\displaystyle f''(x) = \dfrac{-9}{4(3x-7)^{3/2}} -\dfrac{1}{4(71-x)^{3/2}}$ here $f''(x)<0$ so $f'(x)$ is maximum

At maximum $f'(x)= 0$ ,

it leads us to solve $\displaystyle\dfrac{3}{2\sqrt{3x-7}} = \dfrac{1}{2\sqrt{71-x}}$ $\displaystyle \implies \dfrac{3x-17}{9} =71-x \implies x= \dfrac{164}{3}$

Maximum is at $x= \dfrac{164}{3}$ ,

Maximum value $= \sqrt{3.\frac{164}{3}-17 }+\sqrt{71-\frac{164}{3} } = \dfrac{28}{\sqrt{3}} = 16.165..$

Answer is $\boxed{16}$