An Upper Bound

To get an upper bound we use the Root Mean Square - Arithmetic Mean inequality, which states that, for positive real $x_1,\ldots,x_n$ ,

$\qquad\qquad\sqrt{\frac{x_1^2 + x_2^2 \ldots + x_n^2}{n}} \geq \frac{x_1 + x_2 + \ldots + x_n}{n}$

with equality if and only if $x_1 = x_2 = \ldots = x_n$ . We can use this here because all of the square roots are positive and real, as specified in the question.

If we put $x_1 = \sqrt{x+1}, x_2 = \sqrt{2x-3}$ and $x_3 = \sqrt{110-3x}$ , we get that

$\sqrt{\frac{(x+1) + (2x-3) + (110-3x)}{3}} \geq \frac{\sqrt{x+1} + \sqrt{2x-3} + \sqrt{110-3x}}{3}$

or, cancelling the $x$ 's in the left side and swapping the sides,

$\sqrt{x+1} + \sqrt{2x-3} + \sqrt{110-3x} \leq 3\sqrt{\frac{108}{3}} = 3\sqrt{36} = 18$

$\quad$

18 is only actually reached if $\sqrt{x+1} = \sqrt{2x-3} = \sqrt{110-3x}$ . This is, however, impossible, as $x+1 = 2x-3 \Rightarrow x=4$ , but then $110-3(4) \neq (4)+1$ . If you let $x=26$ though, then

$\sqrt{x+1} + \sqrt{2x-3} + \sqrt{110-3x} =$

$\sqrt{27} + \sqrt{49} + \sqrt{32} = 7 + 4\sqrt{2} + 3\sqrt{3} \approx 17.85\ldots > 17$ ,

so 17 is not an upper bound for this equation. Therefore, 18 is the lowest integer upper bound, so $N = 18$ .

First observe that in the common domain [3/2, 110/3] , the function sqrt(x) is concave.

Thus by Jensen's inequality, the LHS <= 3*rt(36) = 18.

Consider x = 19.5. the LHS takes the value of:

sqrt(20.5) + sqrt(36) + sqrt(51.5) > 4 + 6 + 7 = 17.

Thus N = 18

By Cauchy Schwarz Inequality, we have

$(\sqrt{x+1}+\sqrt{2x-3}+\sqrt{110-3x})^2\leq (1+1+1)(x+1+2x-3+110-3x)$

which gives $(\sqrt{x+1}+\sqrt{2x-3}+\sqrt{110-3x})^2\leq 3(108) = 324$ . Thus, $\sqrt{x+1}+\sqrt{2x-3}+\sqrt{110-3x}\leq 18$ and $N = 18$ .

given f(x) = $\sqrt{x+1}+ \sqrt{2x-3}+\sqrt{110-3x}$

and N is actually the maximum value (rounded up integer) of such function.

to find the maximum point, we know that g(x)= $\frac{d f(x)}{dx}=0$

$\frac{1}{2\sqrt{x+1}}+\frac{1}{\sqrt{2x-3}}-\frac{3}{\sqrt{110-3x}}=0$

• this equation is way too complex and i used bisection method (numerical method) to iterate for its solutions,
  g(24)>0 and g(25)<0...... eventually
  

x $\approx$ 24.381 , (...I believe that it's second derivative is < 0 ,so we could have a unique maximum point)

$\therefore f(x_{max})\approx f(24.381)=17.8737 \leq N$

N=18

The common domain is $\frac{3}{2} \leq x \leq \frac{110}{3}$ . Since x is positive, we may apply the Cauchy-Schwarz inequality. $\sqrt{x+1}+\sqrt{2x-3}+\sqrt{110-3x})$ \leq (x+1)+(2x-3)+(110-3x)) (1+1+1)=324).

Let \sqrt{x+1} = a {1}, \sqrt{2x-3} = a {2} and \sqrt{108-3x} = a {3}. By the arithmetic and square means inequality, we have: \sqrt{\frac{a {1}^{2}+a {2}^{2}+a {3}^{2}}{3}} \geq \frac{a {1}+a {2}+a {3}}{3} Using a {1}, a {2} and a {3}, follows that, \frac{\sqrt{x+1}+\sqrt{2x-3}+\sqrt{108-3x}}{3} \leq \sqrt{\frac{\sqrt{(x+1)^{2}+\sqrt{(2x-3)^{2}}+\sqrt{(108-3x)^{2}}}{3}} = \sqrt{\frac{(x+1)+(2x-3)+(108-3x)}{3} \frac{\sqrt{x+1}+\sqrt{2x-3}+\sqrt{108-3x}}{3} \leq \sqrt{\frac{108}{3}} = 6 Then, \sqrt{x+1}+\sqrt{2x-3}+\sqrt{108-3x} \leq 6\times 3 = 18 Verifying the expression \sqrt{x+1}+\sqrt{2x-3}+\sqrt{108-3x} for x=15 \sqrt{24+1}+\sqrt{2\times 24-3}+\sqrt{108-3\tiimes 24} = \sqrt{25}+\sqrt{45}+\sqrt{33} \geq \sqrt{25}+\sqrt{45}+\sqrt{36} = 17 Thus, it is easy to see that 18 is the upper integer boundary of that expression.

The common domain of the left hand side is $1.5 \le x \le \frac{110}{3}$ . Differentiating the LHS, we get $\frac{1}{2\sqrt{x + 1}} + \frac{1}{\sqrt{2x - 3}} - \frac{3}{2\sqrt{110 - 3x}}$ . The roots of the expression stated is between $24$ and $24.5$ by intermediate value theorem. Thus, $5 < \sqrt{x + 1} < \sqrt{25.5}$ , $\sqrt{45} < \sqrt{2x - 3} < \sqrt{46}$ , $\sqrt{35} < \sqrt{110 - 3x} < \sqrt{36.5}$ . One can easily verify that both $5 + \sqrt{45} + \sqrt{35}$ and $\sqrt{25.5} + \sqrt{46} + \sqrt{36.5}$ are between $17$ and $18$ . Thus, $N = 18$ .

conditional: $\frac {3}{2} \leq x \leq \frac {110}{3}$ we have ( \sqrt{x + 1} + \sqrt{2x - 3} + \sqrt{110 - 3x} \leq \sqrt{x + 1 + 2x - 3 + 110 - 3x} = 18

The Cauchy-Schwarz inequality guarantees that for real numbers $u_1$ , $u_2$ , and $u_3$ , $(u_1 + u_2 + u_3 )^2 \le 3(u_1^2 + u_2^2 + u_3^2).$ Hence, we get that $(\sqrt{x+1} + \sqrt{2x-3} + \sqrt{110-3x})^2 \le 3(x+1+2x-3+110-3x) = 324.$ By taking the square root of both sides of this inequality, we obtain the inequality $(\sqrt{x+1} + \sqrt{2x-3} + \sqrt{110-3x}) \le 18.$ To see that this is the smallest integer $N$ , note that $\sqrt{26+1} + \sqrt{2\cdot 26-3} + \sqrt{110-3\cdot 26} = \sqrt{27} + \sqrt{49} + \sqrt{32} > 5 + 7 + 5 = 17.$ Therefore, the smallest such integer $N$ is 18.

The zero of the derivative of the cubic approximation at 25 is potentially very different from the zero of the original function, and the choice of 25 is obviously inspired by ad-hoc numerical arguments, not included...