Hölder's inequality for $\frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}}$

Question

Let $a,b,c \in \mathbb{R}$ and $a+b+c=3$. What is the minimum value of $$\frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}}?$$

I've previously posted an olympiad problem where the solution was dependent on Hölder's inequality. As I'm trying to learn about how to apply this to contest problems surprisingly I couldn't find a whole lot of example problems. It seems that the general version ($\sum_{i=1}^n |x_iy_i| \leqslant (\sum_{i=1}^n|x_i|^p)^{1/p}(\sum_{i=1}^n|y_i|^q)^{1/q}$) of this is not very applicable here.

For this problem that I found it seems that the proposed solution skipped some steps while explaining it. What they had was this:

$$(\frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}})^{1/3}(\frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}})^{1/3}(a + b + c)^{1/3} \geq 1^{1/3} + 1^{1/3} +1^{1/3} = 3.$$

Now it seems that here they have $p=q=3$? How did they come up with the $LHS$ here? It's not clear and there's no explanation for it anywhere...

Answer 1

In your general version, take $p = 3$ and $q = 3/2$.

Take $x_{1, 2, 3} = a^{1/3}, b^{1/3}, c^{1/3}$ and $y_{1, 2, 3} = a^{-1/3}, b^{-1/3}, c^{-1/3}$.

This gives you the minimal value.

Answer 2

The Holder's inequality for two sequences it's the following.

Let $a_1$, $a_2$,..., $a_n$, $b_1$, $b_2$,..., $b_n$, $\alpha$ and $\beta$ be positive numbers. Prove that: $$(a_1+a_2+...+a_n)^{\alpha}(b_1+b_2+...+b_n)^{\beta}\geq$$$$\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}+...+\left(a_n^{\alpha}b_n^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}.$$

For positives $a$, $b$ and $c$ by Holder we obtain: $$\left(\sum_{cyc}\frac{1}{\sqrt{a}}\right)^2\sum_{cyc}a\geq\left(\sum_{cyc}\sqrt[3]{\left(\frac{1}{\sqrt{a}}\right)^2\cdot{a}}\right)^3=(1+1+1)^3=27,$$ which says $$\sum_{cyc}\frac{1}{\sqrt{a}}\geq3.$$ The equality occurs for $a=b=c=1,$ which says that we got a minimal value.