Contest Math: Finding maximum value under restrictions $p,q,r,s$ are non negative real numbers. 
$p^5 + q^5\leq 1$ and $r^5+ s^5 \leq 1$
Find the maximum value of $p^2r^3 + q^2s^3$
I thought of using Holder's Inequality, but couldn't get to any specific maximum value of the expression. 
Of course, using Lagrange Multipliers is a method but not a good one (it's cumbersome)
Could someone please give a detailed solution to the problem? Thanks a lot. 
 A: Using Holder is straightforward,
$$1 \geqslant (p^5+q^5)^{2/5} \cdot (r^5+s^5)^{3/5} \geqslant (p^2r^3+q^2s^3)$$
Equality is possible when $p=q=r=s=\frac1{\sqrt[5]2}$, so that's the maximum.  Any details you need, you should ask for. 
A: Just to share some nice thought on a general case:
Suppose you have $p^a+q^a\leq l_1$ and $r^b+s^b\leq l_2$ with $a,b\in\mathbb{R}_{\geq0}$ and want to maximise $p^fr^g+q^fs^g$ with $f,g\in\mathbb{R}_{\geq 0}$. One of the possible maximal solutions has to consist of only nonnegative numbers, so fractional exponents are ok. Now define $\hat{p}=p^f,\hat{q}=q^f,\hat{r}=r^g,\hat{s}=s^g$. Now you can interpret your quantity to maximise as a scalar product of the vectors $\begin{pmatrix}\hat{p}\\\hat{q}\end{pmatrix}$ and $\begin{pmatrix}\hat{r}\\\hat{s}\end{pmatrix}$, where the conditions turn into $\hat{p}^{\frac{a}{f}}+\hat{q}^{\frac{a}{f}}\leq l_1$ and $\hat{r}^{\frac{b}{g}}+\hat{s}^{\frac{b}{g}}\leq l_1$.
So for the scalar product to be maximal, we have to get the maximal possible modulus for those vectors, but they should also become as collinear as possible. But what if I told you that the maximal modulus for each of the two vectors fulfilling their respective conditions always leads to collinearity? Let's prove it:
Take an arbitrary vector $\begin{pmatrix}x\\y\end{pmatrix}$ fulfilling the condition $x^e+y^e\leq d$ with $e\in\mathbb{R}_{\geq0}$. Then you can find the maximal value for the modulus with Lagrange multipliers rather quickly.
$\nabla_{x,y}(x^2+y^2+\lambda(x^e+y^e-d))=\begin{pmatrix}2x-p\lambda x^{p-2}\\2y-p\lambda y^{p-2}\end{pmatrix}$. Setting this to the null vector and equating the two lambdas leaves us with $\frac{2}{px^{p-3}}=\frac{2}{py^{p-3}}$ leading to $x=y$.
Returning to the original problem, this means the modulus of vectors $\begin{pmatrix}\hat{p}\\\hat{q}\end{pmatrix}$ and $\begin{pmatrix}\hat{r}\\\hat{s}\end{pmatrix}$ is each maximised with respect to their conditions when $\hat{p}=\hat{q}$ and $\hat{r}=\hat{s}$, which are both multiples of $\begin{pmatrix}1\\1\end{pmatrix}$ and therefore collinear. Above equalities also imply $p=q$ and $r=s$ and therefore an easy way to find the maximal values. 
