Let $a,b,c,d$ be non negative real numbers such that $a^5 + b^5 \le 1$ and $c^5 + d^5 \le 1$. Find the maximum possible value of $a^2c^3 + b^2d^3$.

I tried using AM, GM and some other basic inequalities but they were of no use. Need some hints.

  • $\begingroup$ method 1 Hint:Use cauchy schwarz inequality. Do make sure that you invoke the constraint that a,b,c,d must lie between 0 and 1. METHOD 2: Consider the product of (a^2 +b^2+c^2 +d^2)(a^2 +b^3 +c^3 +d^3) $\endgroup$ – Prayas Agrawal Apr 3 '17 at 16:30

It can be done easily with Holder's inequality, with $p=5/2, q=5/3$ and $u=(a^2,b^2), v=(c^3,d^3)$ then:

$$\begin{align}a^2c^3+b^2d^3 &= |(a^2c^3,b^2d^3)|_1\\&\leq |(a^2,b^2)|_p |(c^3,d^3)|_q \\&= (a^5+b^5)^{2/5}(c^5+d^5)^{3/5}\leq 1 \end{align}$$

It now suffices to find a case where it is equal to $1$, which is pretty easy to do.

A more elementary approach is to use AM/GM and get that:

$$\frac{2}{5}a^5+\frac{3}{5}c^5= \frac{1}{5}(a^5+a^5+c^5+c^5+c^5)\geq \sqrt[5]{a^{10}c^{15}} = a^2c^3$$

Similarly: $$\frac{2}{5}b^5+\frac{3}{5}d^5\geq b^2d^3$$ Adding you get:

$$1 = \frac{2}{5}+\frac{3}{5}\geq \frac{2}{5}(a^5+b^5)+\frac{3}{5}(c^5+d^5)\geq a^2c^3+b^2d^3$$

  • 2
    $\begingroup$ +1 For the incredible elegance. But I felt tempted to downvote as you waited till I completed my ugly computations ;) $\endgroup$ – Severin Schraven Apr 3 '17 at 19:10
  • 1
    $\begingroup$ Well, it is certainly elegant, but it requires a result a lot of students don't get to when working on problems like these. Always good to have a "rawer" answer, as well. @SeverinSchraven $\endgroup$ – Thomas Andrews Apr 3 '17 at 19:13
  • $\begingroup$ The more elementary approach was quite tricky to think of. I was also trying some things like that. Thanks. $\endgroup$ – Raghav Chaturvedi Apr 4 '17 at 0:39
  • $\begingroup$ @RaghavChaturvedi I actually worked back from my first answer and saw how Holder's inequality was proved from Young's inequality, which in this case was $xy\leq \frac{2}{5}x^{5/2}+\frac{3}{5}y^{5/3}$. Then I saw the AM-GM proof of Young. $\endgroup$ – Thomas Andrews Apr 4 '17 at 2:02

Let $M=\{(a,b,c,d)\in (\mathbb{R}_{\geq 0})^4 \ : \ a^5+b^5\leq 1, c^5+d^5\leq 1 \}$. Consider

$$ f: \ M \rightarrow \mathbb{R}, \ f(a,b,c,d)= a^2 c^3+b^2 d^3.$$

The derivative is

$$ \nabla f(a,b,c,d)=( 2ac^3, 2bd^2, 3a^2c^3, 3b^2d^2).$$

If the maximum was attained in the interior of $M$, then its gradient vanishes. However,

$$ (0,0,0,0)=\nabla f(a,b,c,d)=( 2ac^3, 2bd^2, 3a^2c^3, 2b^2d)$$

implies that $a=0, b=0$ or $a=0,d=0$ or $c=0,b=0$ or $c=0,d=0$. Clearly, this does not give us a maximum. Hence, the maximum lies on the boundary of $M$.

We are on the boundary, if at least one of $a,b,c,d$ is equal zero (which by the same argument will not give a maximum) or if we are in the following set

$$ \{(a,b,c,d)\in (\mathbb{R}_{\geq 0})^4 \ : \ a^5+b^5= 1 \} \cup \{(a,b,c,d)\in (\mathbb{R}_{\geq 0})^4 \ : \ c^5+d^5= 1 \}.$$

On this set we either have $a=(1-b^5)^{1/5}$ or $c= (1-d^5)^{1/5}$. Thus, we can consider

$$ f_1(b,c,d)=f((1-b^5)^{1/5}, b,c,d) $$


$$ f_2(a,b,d)=f(a,b,(1-d^5)^{1/5},d)$$

and move on with the analysis. Computing the gradient again, we finally can assume $a=(1-b^5)^{1/5}$ and $c= (1-d^5)^{1/5}$. Thus, we have to consider the map

$$ f_3(b,d)= f((1-b^5)^{1/5}, b, (1-d^5)^{1/5}, d) = (1-b^5)^{2/5}\cdot (1-d^5)^{3/5} + b^2\cdot d^3. $$

on the domain $[0,1]\times [0,1]$. For $0<b,d< 1$ we can compute the gradient

\begin{align*} \nabla f_3(b,d)&= \left(\frac{2}{5}\cdot \frac{-5b^4}{(1-b^5)^{3/5}}(1-d^5)^{3/5} +2bd^3, (1-b^5)^{2/5}\cdot \frac{3}{5}\frac{-5d^4}{(1-d^5)^{2/5}}+3b^2 d^2\right) \\ &= \left(2b\cdot \left(\frac{-b^3}{(1-b^5)^{3/5}}(1-d^5)^{3/5} +d^3\right), 3d\cdot \left(-(1-b^5)^{2/5}\cdot \frac{d^3}{(1-d^5)^{2/5}}+b^2 d \right)\right) \\ &= \left( 2b \cdot \left(-b^3\cdot\left(\frac{1-d^5}{1-b^5} \right)^{3/5}+d^3 \right), 3d^2\cdot \left( -d^2 \cdot \left( \frac{1-b^5}{1-d^5} \right)^{2/5} +b^2 \right)\right) \end{align*}

Hence, the gradient vanishes iff (as $0<b,d<1$)

$$ d= b\cdot \left( \frac{1-d^5}{1-b^5} \right)^{1/5} \quad \text{ and } \quad b= d\cdot \left( \frac{1-b^5}{1-d^5} \right)^{1/5}.$$

Assume $b>d$, then (as $0<b,d <1)$ we have

$$ 0< \frac{1-b^5}{1-d^5} <1,$$

which implies

$$ b = d\cdot \left( \frac{1-b^5}{1-d^5} \right)^{1/5} < d. $$

Which is a contradiction. By symmetrie we get $b=d$. In this case we get

$$f_3(b,b)=(1-b^5)+ b^5=1.$$

One easily checks that for $b=0,1$ or $d=0,1$ we still have $f_3(b,d)\leq 1$.

Hence, $\max_{(a,b,c,d)\in M} f(a,b,c,d)=1$.


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