Give me a four, give me a two

If we add two equations on the question, it will be similar to $(x+1)(y+1)(z+1)$ .

We can figure out that $(x+1)(y+1)(z+1)=1331=11^{3}$ .

Because $z+\frac{y}{2}+\frac{x}{4}$ can be changed into $(z+1)+\frac{(y+1)}{2}+\frac{(x+1)}{4}-\frac{7}{4}$ , if we use AM-GM, $(z+1)+\frac{(y+1)}{2}+\frac{(x+1)}{4}-\frac{7}{4} >= 3(\frac{(x+1)(y+1)(z+1)}{8})^{\frac{1}{3}}-\frac{7}{4}=\frac{33}{2}-\frac{7}{4}=\frac{59}{4}$

Hence, $a=59, b=4$ , so the answer is 63.

(More specifically, $z=\frac{9}{2}, y=10, x=21$ , when $(z+1)=\frac{(y+1)}{2}=\frac{(x+1)}{4}$ )

$x(y + 1) + y(z+1) +z(x+1) = 385 \\ \rightarrow xy + x + yz + y + zx + z = 385 \\ \rightarrow xyz + xy + x + yz + y + zx + z = 1330 \\ \rightarrow xyz + xy + x + yz + y + zx + z + 1 = 1331 \\ \rightarrow (x + 1)(y + 1)(z + 1) = 1331 (1)$

By AM - GM inequality, $\\ \frac{4(z + 1) + 2(y + 1) + (x + 1)}{3} \geq \sqrt[3]{4(z + 1) * 2(y + 1) * (x + 1)} \\ 4z + 4 + 2y + 2 + x + 1 \geq 6\sqrt[3]{(z + 1)(y + 1)(x + 1)}$

By (1) $\\ 4z + 2y + x + 7 \geq 66 \\ 4z + 2y + x \geq 59 \\ \frac{4z + 2y + x}{4} \geq \frac{59}{4} \\ z + \frac{y}{2} + \frac{x}{4} \geq \frac{59}{4}$

Therefore minimum value of $z + \frac{y}{2} + \frac{x}{4} = \frac{59}{4} \\ a + b = 59 + 4 = 63$

Firstly, i saw that the minimum value can be easily got by using AM-GM on $\large{z + \frac{y}{2} + \frac{z}{4}}$ but alaas! $\large{xyz}$ is not a perfect cube and i could not get an accurate value. This is what i did then.

On expanding the second equation we get

$\large{x+y+z+xy+yz+zx = 385}$ .

It was indeed familiar expression. Just adding $xyz+1$ to both sides gives us

$\large{1+x+y+z+xy+yz+yz+xyz = 385+945+1 = 1331}$ .

The left hand side can be factored, leaving us with

$(x+1)(y+1)(z+1) = 1331$ .

This is where a little trick has to be performed . Divide both sides by 8, so that we have

$\dfrac{(x+1)}{4} \dfrac{(y+1)}{2} (z+1) = \dfrac{1331}{8}$

$\implies$ $\large{(z+1)(\frac{y}{2} + \frac{1}{2})(\frac{x}{4} + \frac{1}{4})}$ = $\dfrac{11^3}{2^3}$

Now we apply $\text{\large{AM-GM}}$ inequality on the terms $(z+1), (\frac{y}{2} + \frac{1}{2}),(\frac{x}{4} + \frac{1}{4})$

We get

$(z+1)+(\frac{y}{2} + \frac{1}{2})+(\frac{x}{4} + \frac{1}{4}) \geq 3\sqrt[3]{(z+1)(\frac{y}{2} + \frac{1}{2})(\frac{x}{4} + \frac{1}{4})}$

$\implies$ $\large{z + \frac{y}{2} + \frac{z}{4} + \frac{7}{4}} \geq 3\sqrt[3]{(z+1)(\frac{y}{2} + \frac{1}{2})(\frac{x}{4} + \frac{1}{4})} = 3 \times \frac{11}{2} = \frac{33}{2}$

$\implies$ $\Large{z + \frac{y}{2} + \frac{z}{4} \geq \frac{33}{2}- \frac{7}{4}} = \boxed{\dfrac{59}{4}}$ and hence the answer $\large{59+4=\boxed{63}}$

Q.E.D :)

Adding the two equations gives $(x + 1)(y + 1)(z + 1) = 1331$ . By AM-GM,

$z + \frac{y}{2} + \frac{x}{4} = \frac{x + 1}{4} + \frac{y + 1}{2} + (z + 1) - \frac{7}{4} \ge 3\sqrt[3]{\frac{(x + 1)(y + 1)(z + 1)}{8}} - \frac{7}{4} = \frac{33}{2} - \frac{7}{4} = \frac{59}{4}$ . Equality holds iff $x = 21, y = 10, z = 4.5$ . Thus, the minimum possible value of $z + \frac{y}{2} + \frac{x}{4}$ is $\frac{59}{4}$ and the answer is $59 + 4 = \boxed{63}$ .

Combine the first two equations and add one to get $x+xy+yz+z+zx+z+xyz+1=(x+1)(y+1)(z+1)$

$=945+385+1=1331$ .

Rewrite what we want to minimize as $(z+1)-1+\frac{(y+1)}{2}-\frac{1}{2}+\frac{(x+1)}{4}-\frac{1}{4}$

$=(z+1)+\frac{(y+1)}{2}+\frac{(z+1)}{4}-\frac{7}{4}$ .

We know by AM-GM that $(z+1)+\frac{(y+1)}{2}+\frac{(z+1)}{4} \ge 3\cdot \sqrt[3]{(z+1)\cdot\frac{(y+1)}{2}\cdot\frac{(z+1)}{4}}=\frac{33}{2}$ .

Equality is set when $(z+1)=\frac{(y+1)}{2}=\frac{(x+1)}{4}$ and we get $(x,y,z)=(21,10,4.5)$ . We see these numbers satisfy both equations.

$\frac{33}{2}-\frac{7}{4}=\frac{59}{4}$ . $59+4=63$ .

Consider all the divisors of 945. Because it end with a 5, one of the factor must be at least 5. If one of the 3 variable, x ,y or z = 5 we end up with 189. Let's say x=5, then we have :

$yz = 189$

The divisors of 189 are (in pair)

$\begin{array}{l} 1 ; 189\\ 3 ; 63\\ 7 ; 27\\ 9 ; 21 \end{array}$

Then we consider the last equation with those values. z is not divided by anything so giving it the smallest number is the most logical thing to do. All the cases can be written like this :

$\begin{array}{l} 1 + \frac{5}{2} + \frac{{189}}{4} = \frac{{203}}{4}\\ 3 + \frac{5}{2} + \frac{{63}}{4} = \frac{{85}}{4}\\ 5 + \frac{7}{2} + \frac{{27}}{4} = \frac{{61}}{4}\\ 5 + \frac{9}{2} + \frac{{21}}{4} = \frac{{59}}{4} \end{array}$

The smallest value is 59/4 which gives 63 when added together. The answer is 63.

We add up the two equations...

$~$

$\begin{aligned} &~ ~~~& x(y+1)+y(z+1)+z(x+1)+xyz ~&= &945+385 \\ \\ &\Longrightarrow ~~~& xy+x+yz+y+zx+z+xyz ~&= &1330 \\ \\ &\Longrightarrow ~~~& xy+y+zx+z+xyz+yz+x+1~&=&1330+1 \\ \\ &\Longrightarrow ~~~&(x+1)(yz+z+y+1)~&= &1331 \\ \\ &\Longrightarrow ~~~&(x+1)(y+1)(z+1) ~&= &1331 ~~~~~ \cdot \cdot \cdot ~ (1) \\ \end{aligned}$

$~$

We apply the $\textrm{AM-GM inequality}$ on $(1)$ to get...

$~$

$\begin{aligned} &~ ~~~&\frac{(x+1)+2(y+1)+4(z+1)}{3} &\geq &\sqrt[3]{(x+1) \times 2(y+1) \times 4(z+1)} \\ \\ &\Longrightarrow ~~~&\frac{(x+1)+2(y+1)+4(z+1)}{3} &\geq &\sqrt[3]{8(x+1)(y+1)(z+1)} \\ \\ &\Longrightarrow ~~~&(x+1)+2(y+1)+4(z+1) &\geq & 2 \times 3 \sqrt[3]{(x+1)(y+1)(z+1)} \\ \\ &\Longrightarrow ~~~&x+1+2y+2+4z+4 &\geq &6\sqrt[3]{(x+1)(y+1)(z+1)} \\ \\ &\Longrightarrow ~~~&4z+2y+x+7 &\geq &6\sqrt[3]{1331}~~~=6\sqrt[3]{11^3} ~~~~~ \cdot \cdot \cdot ~[\textrm{from}~(1)] \\ \\ &\Longrightarrow ~~~&4z+2y+x+7 &\geq &6 \times 11 ~~~~~=66 \\ \\ &\Longrightarrow ~~~&4z+2y+x &\geq &66-7 ~~~~~= 59 \\ \\ &\Longrightarrow ~~~&\frac{4z+2y+x}{4} &\geq &\frac{59}{4} \\ \\ &\therefore ~~~&z+\frac{y}{2}+\frac{x}{4} &\geq &\frac{59}{4} \\ \\ \end{aligned}$

$~$

Hence, the minimum possible value is: $~~~\textrm{min}(z+\frac{y}{2}+\frac{x}{4})=\frac{a}{b}=\frac{59}{4}$

$~$

Therefore, the required answer is: $~~~a+b=59+4=\fbox{63}$

Adding the given equations, we obtain: $x(y+1)+y(z+1)+z(x+1)+xyz= 945+385$ $\implies (x+1)(y+1)(z+1)= 945+385+1= 1331$

Applying AM-GM inequality on the set $(z, \frac{y+1}{2}, \frac{z+1}{4})$ , we obtain: $z + \frac{y+1}{2} + \frac{x+1}{4} \geq 3 \sqrt[3]{\frac{(x+1)(y+1)(z+1)}{8}}$ $\implies z + \frac{y}{2} + \frac{x}{4} + \frac{1}{2} + \frac{1}{2} \geq 3 \sqrt[3]{\frac{1331}{8}}= 3 \times \frac{11}{2}= \frac {33}{2}$ $\implies z + \frac{y}{2} + \frac{x}{4} \geq \frac{33}{2} - \frac{1}{2} - \frac{1}{4}= \frac{59}{4}$

Equality holds iff $z+1= \frac{y+1}{2}= \frac{x+1}{4}$ . Solving this from the equation $(x+1)(y+1)(z+1)= 1331$ , we obtain $(x,y,z)= (21, 10, \frac{9}{2})$ . Plugging these values in the original equations, we see that they satisfy them. Hence equality holds for the set $(x,y,z)= (21, 10, \frac{9}{2})$ , and the minimum possible value of $z+\frac{y}{2} + \frac{x}{4}$ is $\frac{59}{4}$ . Hence $a= 59$ , $b= 4$ , and $a+b= 59+4= \boxed{63}$ .

The method of Lagrange Multipliers can be used to optimize $f(x,y,z)=z+\frac{y}{2}+\frac{x}{4}$ . Instead of using two constraints, one can simply use a single constraint: one that describes the two simultaneously, i.e. the solution-equation.

$x(y+1)+y(z+1)+z(x+1)=385=(x+1)(y+1)(z+1)-xyz-1 \\ 385=(x+1)(y+1)(z+1)-945-1\\ 1331=(x+1)(y+1)(z+1)$

Letting $\lambda$ be the Lagrange multiplier, then we get the system

$\frac{1}{4}=(x+1)(y+1)\lambda \\ \frac{1}{2}=(z+1)(x+1)\lambda \\ 1=(x+1)(y+1)\lambda$

Dividing the first by the second, and the second by the third gives

$\frac{1}{2} = \frac{y+1}{x+1} \Rightarrow x+1=2(y+1) \\ \frac{1}{2}=\frac{z+1}{y+1} \Rightarrow z+1=\frac{y+1}{2}$

Substituting the expressions in terms of to the constraint $1331=(x+1)(y+1)(z+1)$ yields $y+1=11$ . Hence, we can solve $y=10, x=21$ and $z=\frac{9}{2}$ . Therefore, we get $z+\frac{y}{2}+\frac{x}{4}=\frac{59}{4}$ (One may check whether or not this is the minimum by using the Second-derivative test.)

By AM-GM inequality,

\frac{x}{4} + \frac{1}{4} + \frac{y}{2} + \frac{1}{2} + z+1 >= 3((\frac{x}{4}+\frac{1}{4})(\frac{y}{2}+\frac{1}{2})(z+1))^1/3 = 3((\frac{x+1}{4})(\frac{y+1}{2})(z+1))^1/3 = \frac{3}{8^1/3} * ((x+1)(y+1)(z+1))^1/3 = \frac{3}{2} (xyz + x(y+1)+y(z+1) + z(x+1) +1 )^1/3 = \frac{3}{2} (945 + 385 +1)^1/3 = \frac{3}{2} *1331^1/3 = \frac{3}{2} (11) = \frac{33}{2}

Hence \frac{x}{4} + \frac{y}{2} + z + \frac{7}{4} = \frac{33}{2} and \frac{x}{4} + \frac{y}{2} + z = \frac{59}{4} which means a=59 and b=4. Thus a+b = 63.