# Use Holder's to show $\left( \sum_{\mbox{cyc}} \frac{a}{\sqrt{k(a,b,c)}} \right)^2 \left( \sum_{\mbox{cyc}} ak(a,b,c)\right)\geq (a+b+c)^3$

In one answer to If $abc=1$ so $\sum\limits_{cyc}\frac{a}{\sqrt{a+b^2}}\geq\frac{3}{\sqrt2}$ the statement is made that if $a,b,c$ are positive,

Using Holder's inequality, we have in general $$\left( \frac{a}{\sqrt{X}} +\frac{b}{\sqrt{Y}} + \frac{c}{\sqrt{Z}} \right)^2 \left( aX+bY+cZ \right) \geq (a+b+c)^3$$

(In the context of that problem, there is a constraint that that $abc=1$.)

Obviously, one also needs to know that $X,Y,Z$ are always positive. For my purposes, I only care about the case where $X, Y, Z$ are cyclic-symmetric in $a,b,c$; I will state this by saying that $X=k(a,b,c), Y=k(b,c,a), Z=k(c,a,b)$.

My problem is I don't think I have the correct version of Holder's inequality, or at any rate, the highlighted statement is far from obvious to me. So what I am looking for is:

For $a,b,c > 0$ and $k(a,b,c)>0$ whenever $a,b,c > 0$, use Holder's inequality to prove $$\left( \sum_{\mbox{cyc}} \frac{a}{\sqrt{k(a,b,c)}} \right)^2 \left( \sum_{\mbox{cyc}} ak(a,b,c)\right)\geq (a+b+c)^3$$

and I request that before using Holder's, you present the version of Holder's inequality you are using. The discrete version I have seen is that if $\frac1p+\frac1q = 1$, $$\sum a_kb_k \leq \left( \sum a_k^p \right)^{1/p} \left( \sum b_k^q \right)^{1/q}$$

• I guess you will use this form of Holder: math.stackexchange.com/questions/1073327/… for 3 weights equal to 1/3. – ivan Apr 4 '17 at 16:20
• @Ivan You get a like because the reference shows the right form of Holder's to use. This hint shows me how to prove it. Actually, only two weights are used, $\frac23$ and $\frac13$. – Mark Fischler Apr 4 '17 at 17:03

Let $a_i>0$, $b_i>0$, $\alpha>0$ and $\beta>0$. Prove that: $$\left(a_1+a_2+...+a_n\right)^{\alpha}(b_1+b_2+...+b_n)^{\beta}\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}+...+\left(a_n^{\alpha}b_n^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}$$ This inequality follows immediately from the convexity of $f(x)=x^k$ on $(0,+\infty)$, where $k\geq1$.