If $a$ and $b$ are positive real numbers such that $a^{2015}+b^{2015}=2015$, find the maximum value of $a+b$.

Question

Hint: Just use the power mean inequality $$\left(\frac{a^{2015}+b^{2015}}{2}\right)^{1/2015}\geq\frac{a+b}{2}$$

My Attempt

I was unable to figure out the solution, using the power mean inequality, but I went ahead and used Bergström's inequality. The given condition is $$\frac{a^{2015}}{1^{2014}}+\frac{b^{2015}}{1^{2014}}=2015$$ Using a generalised form of the inequality, we get \begin{align*} 2015&\geq\frac{[a+b]^{2015}}{[1+1]^{2014}}\\ (2)^{2014/2015}(2015)^{1/2015)}&\geq a+b\\ 2.006&\geq a+b\\ \end{align*} I plugged in the $2.006$ value and it worked. However, I'd like to use Hölder's inequality or the hint to solve this. Would you have any tips?

Answer 1

Using the given hint: \begin{align*} \left(\frac{a^{2015}+b^{2015}}{2}\right)^{\frac{1}{2015}} &\geq \frac{a+b}{2} \\ 2 \cdot \left(\frac{2015}{2}\right)^{\frac{1}{2015}} &\geq a+b \quad (\because a^{2015}+b^{2015}=2015) \\ a+b &\leq 2.0068 \end{align*}

Answer 2

For completeness, here is a proof of the inequality via Hölder’s inequality as OP asked for one:

First, note $\frac{1}{2015}+\frac{2014}{2015}=\frac{1}{2015}+\frac{1}{\frac{2015}{2014}}=1$. Then, by Hölder,

\begin{align*} a\cdot1+b\cdot1&\le\left(a^{2015}+b^{2015}\right)^{\frac{1}{2015}}\left(1^ {\frac{2015}{2014}}+1^{\frac{2015}{2014}}\right)^{\frac{2014}{2015}}\\ a+b&\leq2015^{\frac{1}{2015}}\cdot 2^{\frac{2014}{2015}}\approx2.006 \end{align*}

Of course, using the Power Mean inequality would be more direct and simple.

Answer 3

For constrained optimization problem, you look at the stationary points and boundary points. The target function is $F(a,b) = a + b$, the constraint function is $G(a,b) = a^{2015} + b^{2015} - 2015 = 0$ and the boundary is that for the region $a > 0$ and $b > 0$.

The stationary points are where $dF = 0$ and $dG = 0$: $$dF = da + db = 0, \quad dG = 2015 \left(a^{2014} da + b^{2014} db\right) = 0.$$ Substituting for $db$ from the first into the second yields $$2015 \left(a^{2014} - b^{2014}\right) da = 0,$$ thus $$a^{2014} = b^{2014},$$ from which follows $$a = b.$$ Substituting into $G(a,b) = 0$ yields $$a^{2015} = \frac{2015}{2} = b^{2015}\quad→\quad a = \left(\frac{2015}{2}\right)^{1/2015} = b\quad→\quad a + b = 2\left(\frac{2015}{2}\right)^{1/2015}.$$ For the boundary, $$a = 0\quad→\quad b = 2015^{1/2015}\quad→\quad a + b = 2015^{1/2015},$$ and $$b = 0\quad→\quad a = 2015^{1/2015}\quad→\quad a + b = 2015^{1/2015}.$$

The boundary would actually be excluded by the problem statement, so if either of these were larger than the value of $F(a,b) = a + b$ at the stationary point (they're not), then there'd be no maximum, only a supremum. Instead, they're both smaller and equal to each other, so that means there's no minimum, but only an infinimum - $2015^{1/2015}$ - and the stationary point value $$2\left(\frac{2015}{2}\right)^{1/2015}$$ is the maximum.

The hint is not needed.