Given three non-negative (as pointed out by Calvin Lin) real numbers $x+y+z = 3$, find the maximum value of $\sqrt{2x + 13} + \sqrt[3]{3y+5} + \sqrt[4]{8z+12}$.

(Source : Singapore Math Olympiad 2012, Senior section, Round 1, question 29).

I tried using the fact that $2x +13, 8z + 12\ge 0$ to deduce that $y \le 11$, but I couldn't continue from there on. The answer should be an integer, since only integer answers were allowed.

The competition was designed for 15/16 year olds. A simple yet elegant solution would be nice.

For reference of the original problem, enter image description here

  • 1
    $\begingroup$ If you allow for negative numbers, i don't think there is a maximum. $\endgroup$ – Calvin Lin Apr 24 '14 at 4:38
  • $\begingroup$ I was doubtful about that condition too, but it seems that the problem allowed it anyway. $\endgroup$ – Yiyuan Lee Apr 24 '14 at 4:39

\begin{align*}&\sqrt{\dfrac{2x+13}{4}}\cdot\sqrt{4}+\sqrt[3]{\dfrac{3y+5}{4}}\cdot\sqrt[3]{2}\cdot\sqrt[3]{2}+\sqrt[4]{\dfrac{8z+12}{8}}\cdot\sqrt[4]{2}\cdot\sqrt[4]{2}\cdot\sqrt[4]{2}\\ &\le\dfrac{\dfrac{2x+13}{4}+4}{2}+\dfrac{\dfrac{3y+5}{4}+2+2}{3}+\dfrac{\dfrac{8z+12}{8}+2+2+2}{4}\\ &=\dfrac{1}{4}(x+y+z)+\dfrac{29}{4}\\ &=8 \end{align*}

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    $\begingroup$ how do you know $3y+5 \geq 0 $ ? $\endgroup$ – user139708 Apr 24 '14 at 4:08
  • $\begingroup$ $x,y,z$ can be negative, that's the twist to the problem. $\endgroup$ – Yiyuan Lee Apr 24 '14 at 4:11
  • $\begingroup$ @Lemur,when $\sqrt{2x+13}+\sqrt[3]{3y+15}+\sqrt[4]{8z+12}$ is maximum,we must $3y+15\ge 0$,can maximum $\endgroup$ – math110 Apr 24 '14 at 4:15
  • $\begingroup$ you have to show $3y+5 \geq 0 $ in order to use the AM-GM inequality in the expression involving the cubic root $\endgroup$ – user139708 Apr 24 '14 at 4:17
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    $\begingroup$ Take $ x = 1003, y = -1000, z = 0$, you get 32.38 $\endgroup$ – Calvin Lin Apr 24 '14 at 4:42

If $y$ is allowed to be negative, then there is no maximum.

Take $ x =k - \frac{13}{2}$, $ y = -k - \frac{5}{3}$, $ z= 3 + \frac{13}{2} + \frac{5}{3}$

The important part of the value looks like $ \sqrt{2k} + \sqrt[3]{-3k} + C$, which tends to infinity as $k\rightarrow \infty$.

Otherwise, if $y$ is positive, see Math110's solution.

  • $\begingroup$ You are right, thanks! $\endgroup$ – Yiyuan Lee Apr 24 '14 at 4:52

math110's AM-GM solution is no doubt (to me at least) what the author's wanted. But a less inspired approach is that with Lagrange Multipliers, you need to solve the system of equations:

$$\left\{\begin{aligned} x+y+z&=3\\ \frac{1}{\sqrt{2x+13}}&=\lambda\\ \frac{1}{\sqrt[3]{(3y+5)^2}}&=\lambda\\ \frac{2}{\sqrt[4]{(8z+12)^3}}&=\lambda\\ \end{aligned}\right.$$

The second and third imply that $\sqrt{2x+13}=\sqrt[3]{(3y+5)^2}$, so $(2x+13)^3=(3y+5)^4$.

Similarly the second and fourth imply that $2\sqrt{2x+13}=\sqrt[4]{(8z+12)^3}$, so $16(2x+13)^2=(8z+12)^3$.

Now you can forget about $\lambda$ and work with the system $$\left\{\begin{aligned} x+y+z&=3\\ (2x+13)^3&=(3y+5)^4\\ 16(2x+13)^2&=(8z+12)^3\\ \end{aligned}\right.$$

Eliminating $z$: $$\left\{\begin{aligned} (2x+13)^3&=(3y+5)^4\\ (2x+13)^2&=4(9-2x-2y)^3\\ \end{aligned}\right.$$

If there are any solutions where the sides of these equations are integers, we would have to have some integer to the $12$th on both sides of the first equation. Trying something nice and small like $2^{12}$ implies that $x$ would have to be $1.5$ and $y$ would have to be $1$. And this is also a solution to the second equation. So one solution is $(x,y,z)=(1.5,1,0.5)$, which yields a value of $8$. Maybe the relative simplicity of the the curves in this last system can show that there are no other solutions:

It's a little tedious, but we can solve for $y$ in each equation, and the difference of the two functions of $y$ is $$\frac{(2x+13)^{3/4}-5}{3}-\left(\frac{\left(\frac{(2x+13)^2}{4}\right)^{1/3}-9+2x}{2}\right)$$ whose derivative is $$\frac{1}{2(2x+13)^{1/4}}-\frac{2^{1/3}}{3(2x+13)^{1/3}}-1$$ For positve $x$, this is always negative. So the curves from the last system above can only cross once in the domain where $x$ is positive, and we already found where they do.

And then it remains to show that this yields a local maximum value, not a minimum value or a degenerate solution to the Lagrange multiplier problem.

If there are constraints on $x$, $y$, $z$ being positive, then the boundaries of the planes $x=0$ (subject to $y+z=3$), $y=0$ (subject to $x+z=3$), and $z=0$ (subject to $x+y=3$) need to be separately checked, followed by the boundaries of those boundaries: $(3,0,0)$, $(0,3,0)$, and $(0,0,3)$.

  • $\begingroup$ You need to check boundary conditions. All that you have shown is a local maximum exists. See my solution for why the boundary condition gives you no global maximum. $\endgroup$ – Calvin Lin Apr 24 '14 at 4:43
  • $\begingroup$ @Calvin, Thanks - still editing. I write these things in mini posts. $\endgroup$ – alex.jordan Apr 24 '14 at 4:46
  • $\begingroup$ Sorry, i tend to jump at calculus solutions which don't look at boundary conditions (or second order conditions). It's a common mistake that is often overlooked. If you edit, I can undo my down vote. $\endgroup$ – Calvin Lin Apr 24 '14 at 4:49
  • $\begingroup$ @Calvin I suspect OP's original problem asked for positive $x,y,z$. $\endgroup$ – alex.jordan Apr 24 '14 at 4:52
  • $\begingroup$ I looked up the reference, no indication of positive - cl.ly/image/2m3v170q1n2W. I could check with the team and see if it's an error .... $\endgroup$ – Calvin Lin Apr 24 '14 at 4:52

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