Convexity, yeah!

According to another thread I saw, the right side squared item should be a minus (between the two square rooted terms). And in this case, there is no equality...

zapi2007 wrote: According to another thread I saw, the right side squared item should be a minus (between the two square rooted terms). And in this case, there is no equality... Thanks! It's fixed now.

This is killed by convexity, but I would just like to mention another way.

could U explain the other way plz?! :

Sorry, I was not being very careful when I posted that... it doesn't work directly here. You would probably have to do some smoothing on $e$ after applying the linked inequality on $a,b,c,d$, so just use convexity instead.

What do you guys mean by "convexity"? Do you mean Jensen's inequality? Sorry I'm a noob at inequalities...

Treating it as a function in the variable $a$, the LHS is convex (i.e. second derivative is nonnegative). Similarly, it is convex in $b,c,d,e$, so because the variables are "independent" and bounded, the maximum of the LHS occurs on the boundary, when each variable is either $p$ or $q$.

see also: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=367263

Why doesn't AM-GM work? $\textbf{Lemma 1:} \frac{a+b +c +d +e+ \frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e}}{2}\ge 5$. Proof: By AM-GM, $a+\frac{1}{a}\ge 2$. Since this is similar for the other variables, we have the minimum of $\frac{a+b +c +d +e+ \frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e}}{2}$ is $5$. $\blacksquare$ $\textbf{Lemma 2:} 25 \ge (a+b+c+d+e)(\frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e})$. Proof: We apply two variable AM-GM to $a+b+c+d+e$ and $\frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e}$ to get $$(\frac{a+b +c +d +e+ \frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e}}{2})\ge \sqrt{(a+b+c+d+e)(\frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e})}.$$By Lemma 1, we have $$(\frac{a+b +c +d +e+ \frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e}}{2})\ge 5 \ge \sqrt{(a+b+c+d+e)(\frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e})}.$$ We consider the RHS inequality. Squaring both sides is valid because they have the same sign, so we have $25 \ge (a+b+c+d+e)(\frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e})$. $\blacksquare$ $\textbf{Lemma 3:} 25+6(\sqrt{\frac {p}{q}}-\sqrt{\frac{q}{p}})^2 \ge 25.$ Proof: Subtracting $25$ from both sides then dividing by $6$ gives $(\sqrt{\frac {p}{q}}-\sqrt{\frac{q}{p}})^2\ge 0$. This is intuitively clear by the trivial inequality. $\blacksquare$ By Lemmas 1 and 2 along with the transitive property of inequality, we have \[(a+b +c +d +e)\left(\frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e}\right) \le 25 + 6\left(\sqrt{\frac {p}{q}} - \sqrt {\frac{q}{p}}\right)^2\]$\blacksquare$

sketchcomedyrules wrote: $\textbf{Lemma 2:} 25 \ge (a+b+c+d+e)(\frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e})$. This is false. Take $a=1,b=c=d=e=2$ We have $25 \le (a+b+c+d+e)(\frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e})$. By C.S

Ah so it doesnt work. What is the fallacy with amgm there then?

sketchcomedyrules wrote: Ah so it doesnt work. What is the fallacy with amgm there then? Just because $x\geq y$ and $x\geq z$ doesn't mean $x\geq y \geq z$. Similarly, even though we know $ \frac{a+b +c +d +e+ \frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e}}{2}\ge 5$ and $(\frac{a+b +c +d +e+ \frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e}}{2})\ge \sqrt{(a+b+c+d+e)(\frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e})}$, all we know about $ \sqrt{(a+b+c+d+e)(\frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e})}$ and $5$ is that they are both less than the same thing, so we need to do more work to determine if one is always less than the other or not.

Convexity/concavity also helps.

I wonder the problem is from USA MO

Everyone here is saying to use convexity/concavity, but nobody has posted an actual solution to this. I don't understand how the bound on a,b,c,d,e in [p,q] helps, so could somebody give me a hint to get started other than just saying convexity? I don't understand what function to define to use Jensen's, nor do I know any derivative calculus, so if there is a different way, please help.

Would this help? https://easylearn.baidu.com/edu-page/tiangong/questiondetail?id=1727322693010128965&fr=search