Proving inequality contrary to Cauchy Schwarz using additional term Let $0 \lt p\le a,b,c,d,e\le q$. Prove that $$(a+b+c+d+e)\left(\frac 1a + \frac 1b +\frac 1c+\frac 1d+\frac 1e\right)\le 25+6\left(\sqrt {\frac pq}-\sqrt {\frac qp}\right)^2 $$ In the solution they use the fact that $f(a,b,c,d,e)$ is a convex on all variables. Hence the maxima occurs when the variables are on one of the 32 vertices of a 5-cube implying that at the maxima some of the variables are p's and the others are q's. Could someone give me an easier proof without using that concept, as i don't understand that concept... or you could share a link explaining the concept in detail.
 A: Let's consider the variable $a$ for a moment. If we think of fixing everything else and changing $a$, the left hand side looks like
$$ f(a) = (a + x)\left(\frac{1}{a} + y\right) = 1 + \frac{x}{a} + ay + xy$$
so 
$$ f'(a) = y - \frac{x}{a^2}$$
and
$$ f''(a) = \frac{2x}{a^3} > 0,$$
which means that the function is convex in $a$. What this means is that it sort of looks like some part of the parabola $y = x^2$ in that it has an upwards U sort of shape. And if you think about an upwards U, or any subsection of an upwards U, you'll see that the highest it gets is at one of the endpoints of the subsection. Although that's heuristic, it's true in general for convex functions.
The second derivative is positive in each of the variables, independent of the values of the other variables (as long as each is positive, which they are by assumption). The computation is exactly the same. As it's convex in each of the variables, this reduces your consideration to the $32$ possibilities at the endpoints, as you'd mentioned.
Does that make sense?
