Sequential roots

By the AM-GM inequality , we have:

$\begin{aligned} \frac{a + 4b} 2 &\ge \sqrt{a\cdot 4b}\implies \sqrt{ab} \le \frac{a+4b} 4,\\ \frac{a+4b+16c} 3 & \ge \sqrt[3]{a\cdot 4b\cdot 16c} \implies \sqrt[3]{abc} \le \frac{a+4b+16c}{12}.\end{aligned}$

Adding these two equations together and an $a$ to both sides, we get:

$a + \sqrt{ab} + \sqrt[3]{abc} \le \frac 4 3 (a+b+c) = \frac{400} 3.$

Voila!

From the second equation, equality case holds at $a=4b=16c.$

For some inspiration on how to discover this approach, check out Anqi Li's solution.

Since this is an inequality question and there are square (and cube) roots in the question, one natural method to use is AM-GM .

My first instinct was to use AM-GM directly on the roots: $a + \sqrt{ab} + \sqrt[3]{abc}\le a + \frac{a+b}{2}+ \frac{a+b+c}{3}.$ This means that I want to minimize $b$ and $c.$ However, they cannot be 0 since then the product would be 0 and the expression is minimized, instead of maximized and so we are stuck.

Conclusion: we cannot just use AM-GM directly.

Next, I tried something else:

By AM-GM, we have $\sqrt{ab} \le \frac{a+b}{2}$ , which is equivalent to $2 \sqrt{ab} \le a+b$ and similarly $3\sqrt[3]{abc} \le a + b + c.$ With this in mind, we try to write $\sqrt{ab}$ and $\sqrt[3]{abc}$ in the given expression as $2\sqrt{\frac{a}{x} \times\frac{b}{y}}$ and $3\sqrt[3]{\frac{a}{k} \times \frac{b}{m} \times \frac{c}{n}}$ , respectively (where $x \times y = 4$ and $k \times m \times n = 27$ ), so that from AM-GM we have that $a + 2 \sqrt{\frac{a}{x} \times \frac{b}{y}} + 3 \sqrt[3]{ \frac{a}{k} \times \frac{b}{m} \times \frac{c}{n}} \le a + \frac{a}{x} + \frac{b}{y}+ \frac{a}{k} + \frac{b}{m}+ \frac{c}{n}.$

The reason we do this is because we are given the value of $a + b + c$ , so if we can choose values of $x, y, k, m, n$ such that the RHS is a multiple of $a + b + c$ , then we can find a maximum for the given expression. Then we need the coefficients of $a, b, c$ to be equal to each other: in other words, $1 + \frac{1}{x} + \frac{1}{k} = \frac{1}{y} + \frac{1}{m}= \frac{1}{n} .$

Given that $x \times y = 4$ and $k \times m \times n =27,$ to make life easier, we can set $x = 4$ and $y = 1$ so that $1 = 1/y$ , which changes the equation to $\frac{1}{4} + \frac{1}{k} = \frac{1}{m} = \frac{1}{n} - 1.$

To help, in order for the maximum to even be achievable, both of the AM-GM inequalities need to hold, meaning that $\frac{a}{x} = \frac{b}{y}$ and $\frac{a}{k} = \frac{b}{m} = \frac{c}{n}$ . But this means that $\frac{x}{y} = \frac{a}{b} = \frac{k}{m}$ , so we need to have $\frac{k}{m} = \frac{4}{1} = 4$ , or $k = 4m$ . Then since we also need $\frac{1}{4} + \frac{1}{k} = \frac{1}{m}$ , we can see that $\frac{1}{4} = \frac{3}{4m}$ , or $m = 3$ . Then this gives $k = 12$ and $n = \frac{4}{3}$ .

Hence we have (due to AM - GM): $a + 2 \sqrt{ \frac{a}{4} \times b} + 3 \sqrt[3]{\frac{a}{12} \times \frac{b}{3} \times \frac{4c}{3}} \le a + \frac{a}{4} + \frac{a}{12} + \frac{b}{3}+ \frac{4c}{3}$ = $\frac{4}{3} \times (a+b+c),$ with equality achieved when $\frac{a}{4} = b$ and $\frac{a}{12} = \frac{b}{3} = \frac{4c}{3}$ , which happens when $a = \frac{160}{21}, b = \frac{40}{21}$ and $c = \frac{10}{21}$ .

Putting this into the context of the question, we get the maximum is $\frac{m}{n}= \frac{4}{3} \times (a + b + c) = \frac{4}{3} \times 100 = \frac{400}{3} .$ Hence, $m+n = 403.$

Although this can be greatly simplified, I think it is better to share my thoughts and motivations.

Lemma

Let $a,b,c$ be the positive real numbers . Then $a+\sqrt{ab}+\sqrt[3]{abc} \le \frac{4(a+b+c)}{3}$ holds . Equality when $a=4b=16c$

Proof

$a+\sqrt{ab}+\sqrt[3]{abc}=a+2\sqrt{\frac{a}{4}.b}+3\sqrt[3]{\frac{a}{12}.\frac{b}{3}.\frac{4c}{3}}$ Using AM-GM inequality , $a+2\sqrt{\frac{a}{4}.b}+3\sqrt[3]{\frac{a}{12}.\frac{b}{3}.\frac{4c}{3}} \le a+\frac{a}{4}+b+\frac{a}{12}+\frac{b}{3}+\frac{4c}{3}$ Which is equal to $\frac{4}{3}(a+b+c)$ . which completes the proof of lemma .. inequality holds when $\frac{a}{12} = \frac{b}{3} = \frac{4c}{3} \implies a=4b=16c \Box$ .

So using the lemma , we have

$a+\sqrt{ab}+\sqrt[3]{abc} \le \frac{4(100)}{3} = \frac{400}{3}$ So maximum value is $\frac{400}{3}$ Thus $m+n=403 \Box$

Let $a+b+c=S$ . Note that by AM-GM inequality,

$\sqrt{ab} \leq \frac{\frac{1}{2}a+2b}{2}$

$\sqrt[3]{abc} \leq \frac{\frac{1}{4}a+b+4c}{3}$

Combining these, we have $a+\sqrt{ab}+\sqrt[3]{abc} \leq \frac{4}{3}S$ . Further, we can verify that equality occurs at $(a,b,c)=(\frac{16}{21}S,\frac{4}{21}S,\frac{1}{21}S)$ . Hence, the answer is $\frac{4}{3}S=\boxed{\frac{400}{3}}$ .

By using Cauchy inequality, we have:

$\frac{a}{2}+2b \geq 2 \sqrt{ab}$

$\frac{a}{4}+b+4c \geq 3 \sqrt[3]{abc}$ .

Therefore, $a+\sqrt{ab}+\sqrt[3]{abc} \leq \frac{4}{3} (a+b+c)$ .

Therefore, $a+\sqrt{ab}+\sqrt[3]{abc}$ max = $\frac{400}{3}$ when $\frac{a}{4}=b=4c=\frac{400}{21}$ .

The answer is: 400+3=403.

This solution was possible only after help from Sreejato Bhattacharya. Since $a,b,c$ are positive reals, we have

$\dfrac{a + 4b}{2} \geq 2\sqrt{ab}$ $\implies$ $\dfrac{a + 4b}{4} \geq \sqrt{ab}$

Again, $\dfrac{a + 4b + 16c}{3} \geq 4\sqrt[3]{abc}$ $\implies$ $\dfrac{a + 4b + 16c}{12} \geq \sqrt[3]{abc}$

Adding, we get,

$\dfrac{a + 4b}{4} + \dfrac{a + 4b + 16c}{12} \geq \sqrt{ab} + \sqrt[3]{abc}$

Simplification of L.H.S leads to

$\dfrac{4a + 16b + 16c}{12} \geq \sqrt{ab} + \sqrt[3]{abc}$ . Adding a to both sides

$a + \dfrac{4a + 16b + 16c}{12} \geq a + \sqrt{ab} + \sqrt[3]{abc}$

$\implies$ $\dfrac{16(a+b+c)}{12} \geq a + \sqrt{ab} + \sqrt[3]{abc}$

Thus, $a + \sqrt{ab} + \sqrt[3]{abc} \leq \dfrac{4}{3} \times 100 = \boxed{\dfrac{400}{3}}$

Applying the AM-GM inequality, we have:

$a+\sqrt{ab}+\sqrt[3]{abc}$

$\displaystyle =a+\frac{1}{2}{a.4b}+\frac{1}{4}\sqrt[3]{a.4b.16c}$

$\displaystyle \leq a+\frac{a+4b}{4}+\frac{a+4b+16c}{12}$

$\displaystyle =\frac{4}{3}(a+b+c)=\frac{400}{3}$

The equality occurs iff $\displaystyle a=\frac{1600}{21},b=\frac{400}{21},c=\frac{100}{21}$

So $m+n=\boxed{403}$

Let $x = a/16$ , $y = b/4$ , and $z = c$ , so $a = 16x$ , $b = 4y$ , and $c = z$ . Then the given condition $a + b + c = 100$ becomes $16x + 4y + z = 100$ . We want to maximize $$a + \sqrt{ab} + \sqrt[3]{abc} = 16x + 8 \sqrt{xy} + 4 \sqrt[3]{xyz}.$$

By the AM-GM inequality, $\sqrt{xy} \le (x + y)/2$ and $\sqrt[3]{xyz} \le (x + y + z)/3$ , so

$\begin{aligned} 16x + 8 \sqrt{xy} + 4 \sqrt[3]{xyz} &\le 16x + 4(x + y) + \frac{4}{3} (x + y + z) \\ &= \frac{4}{3} (16x + 4y + z) \\ &= \frac{400}{3}. \end{aligned}$

Equality occurs when $x = y = z = 100/21$ , so the maximum value is 400/3.

Because $a,b,c>0$ , by A.M.-G.M. inequality we get

$a+\sqrt{ab}+\sqrt[3]{abc} = a+\frac{\sqrt{a \cdot 4b}}{2}+\frac{\sqrt[3]{a \cdot 4b \cdot 16c}}{4}$ .

$\le a+\frac{1}{2}\cdot\frac{a+4b}{2}+\frac{1}{4}\cdot\frac{a+4b+16c}{3}$

$\le \frac{16a+16b+16c}{12}$

$\le \frac{400}{3}$

Therefore, since $\gcd(400,3)=1$ , the answer is $400+3=403$ .

Apply the AM-GM inequality:

$\sqrt { ab } =\sqrt { \left( \frac { a }{ x } \right) (bx) } \le \frac { \frac { a }{ x } +bx }{ 2 } =\frac { 1 }{ 2x } a+\frac { x }{ 2 } b\\ \\ \sqrt [ 3 ]{ abc } =\sqrt [ 3 ]{ \left( \frac { a }{ y } \right) \left( \frac { b }{ z } \right) (cyz) } \le \frac { \frac { a }{ y } +\frac { b }{ z } +cyz }{ 3 } =\frac { 1 }{ 3y } a+\frac { 1 }{ 3z } b+\frac { yz }{ 3 } c\\ \\ \Rightarrow \quad a+\sqrt { ab } +\sqrt [ 3 ]{ abc } \le \left( 1+\frac { 1 }{ 2x } +\frac { 1 }{ 3y } \right) a+\left( \frac { x }{ 2 } +\frac { 1 }{ 3z } \right) b+\frac { yz }{ 3 } c\quad \quad (*)\\$ where $x,y,z>0$

However, because the only given data we have is $a+b+c=100$ , our goal is to find a positive constant $k$ such that $a+\sqrt { ab } +\sqrt [ 3 ]{ abc } \le \quad k(a+b+c)\quad =\quad ka+kb+kc$ .

So, try to set $1+\frac { 1 }{ 2x } +\frac { 1 }{ 3y } =\frac { x }{ 2 } +\frac { 1 }{ 3z } =\frac { yz }{ 3 }$ . This yields a very beautiful and comfortable solution: $x=2,y=4,z=1$ .

Plug in again to the inequality $(*)$ :

$a+\sqrt { \left( \frac { a }{ 2 } \right) (2b) } +\quad \sqrt [ 3 ]{ \left( \frac { a }{ 4 } \right) \left( \frac { b }{ 1 } \right) (4c) } =a+\sqrt { ab } +\sqrt [ 3 ]{ abc } \le \frac { 4 }{ 3 } a+\frac { 4 }{ 3 } b+\frac { 4 }{ 3 } c=\frac { 400 }{ 3 }$ .

Equality happen when: $\frac { a }{ 2 } =2b\quad and\quad \frac { a }{ 4 } =b=4c\quad and\quad a+b+c=100\\ \Rightarrow \quad a=\frac { 1600 }{ 21 } ;b=\frac { 400 }{ 21 } ;c=\frac { 100 }{ 21 }$

This is my solution!