Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For $a\ge \frac{1}{8}$, we define, $$g(a)=\sqrt[\Large3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[\Large3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}$$

Find the maximum value of $g(a)$.

I came across this question in a Math Olympiad Competition and I am not sure how to solve it. Can anyone help? Thanks.

share|cite|improve this question
I've made an edeit to your post. I believe this is what you ment? If not correct me.... – gebruiker Jun 6 '14 at 13:52
Due to the complexity of the equation(hard to type), I decided to use an image instead. – snivysteel Jun 6 '14 at 13:53
Very smart. I'll try to get is fixed based on the image... – gebruiker Jun 6 '14 at 13:53
I fixed it, the image was a little hard to decipher, is it ok now? – Pranav Arora Jun 6 '14 at 13:55
Yeah, thanks for your help! – snivysteel Jun 6 '14 at 13:58
up vote 3 down vote accepted

In general we cannot rewrite a cubic radical in a simple way. However both radicals in $g(a)$ are of a special form that can be denested, because we can find two cubic powers $X^3,Y^3$ such that $X, Y $ are quadratic conjugate irrationals and

\begin{equation*} a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}=X^{3},\qquad a-\frac{a+1}{3}\sqrt{\frac{ 8a-1}{3}}=Y^{3}. \end{equation*}

If we write $x=\frac{8a-1}{3}\geq 0$, then $a=\frac{3x+1}{8}$ and $\frac{a+1 }{3}=\frac{x+3}{8}$. Consequently, the first radicand becomes

\begin{eqnarray*} a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}} &=&\frac{3x+1+\left( x+3\right) \sqrt{x} }{8}=\frac{1+3\sqrt{x}+3x+\left( \sqrt{x}\right) ^{3}}{8} \\ &=&\frac{\left( 1+\sqrt{x}\right) ^{3}}{2^{3}}=X^{3},\qquad X=\frac{ 1+\sqrt{x} }{2}=\frac {1}{2}+\frac {1}{2}\sqrt {\frac {8a-1}{3}}, \end{eqnarray*}

where we used the binomial theorem for the cubic power

\begin{equation*} \left( 1+c\right) ^{3}=1+3c+3c^{2}+c^{3}, \end{equation*}

with $c=\sqrt{x}$:

\begin{eqnarray*} \left( 1+\sqrt{x}\right) ^{3} &=&\left( 1+x^{1/2}\right) ^{3}=1+3x^{1/2}+3x+x^{3/2} \\ &=&1+3\sqrt{x}+3x+\left( \sqrt{x}\right) ^{3}. \end{eqnarray*}

Similarly, using the binomial theorem for $(1-c)^3=(1-\sqrt{x})^3$, the second radical becomes

\begin{eqnarray*} a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}} &=&\frac{3x+1-\left( x+3\right) \sqrt{x} }{8}=\frac{ 1-3\sqrt{x}+3x-\left( \sqrt{x}\right) ^{3}}{2^{3}} \\ &=&\frac{\left( 1-\sqrt{x}\right) ^{3}}{2^{3}}=Y^{3},\qquad Y=\frac{ 1-\sqrt{x} }{2}=\frac {1}{2}-\frac {1}{2}\sqrt {\frac {8a-1}{3}}. \end{eqnarray*}

Now we find easily that for $a\geq 1/8$ the function $g(a)$ is constant

\begin{equation*} g(a)=\sqrt[3]{X^{3}}+\sqrt[3]{Y^{3}}=X+Y=\frac{ 1+\sqrt{x} }{2} +\frac{ 1-\sqrt{x} }{2} =1. \end{equation*}

Hence $\max_{a\geq 1/8}g(a)=1.$

share|cite|improve this answer

Hint: Let $u =\sqrt{\frac{8a-1}{3}}$ which implies $a = \frac{3u^2+1}{8}$. Then you have to find the max of \begin{align*} \sqrt[3]{\frac{3u^2+1+u^3+3u}{8}}+\sqrt[3]{\frac{3u^2+1-u^3-3u}{8}}. \end{align*} Observe the expansions of $(1+u)^3$ and $(1-u)^3$.

share|cite|improve this answer

Note the $$(a+b)^3=a^3+b^3+3ab(a+b)$$ so $$g^3(a)=2a+3\sqrt[3]{a^2-\dfrac{(a+1)^2}{9}\cdot\dfrac{8a-1}{3}}\cdot g(a)$$

share|cite|improve this answer
How would you continue? – Calvin Lin Jun 6 '14 at 14:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.