Hausdorff-Young inequality

Question

Let $1<p\leq2\leq q \leq \infty$ and let:

$$ \frac{1}{p} + \frac{1}{q}=1 $$

prove that for all finite Abel groups and all functions $f:\mathbb{A}\rightarrow \mathbb{C}$ Hausdorff-Young inequality holds:

$$ \left( \frac{1}{|\mathbb{A}|}\sum_{\xi\in\mathbb{A}}|\tilde{f}(\xi)|\right)^\frac{1}{q}\leq \left( \sum_{\chi\in\mathbb{A}}|f(\chi)| \right)^\frac{1}{p} $$

Then prove following inequalities, $a,b\in \mathbb{C}$:

$$(\frac{1}{2}|a+b|^q+\frac{1}{2}|a-b|^q)^{1/q}\leq (|a|^p+|b|^p)^{1/p}$$

Also for $\omega=e^{i \frac{2\pi}{3}}$ and $a,b,c \in \mathbb{C}$ prove that:

$$ |a+b+c|^3 + |a+\omega b+\omega^2c|^3 + |a+\omega^2b+\omega^3 c|^3 \geq 3(|a|^3+|b|^3+|c|^3) $$

I have tried to find some reasonable literature regarding Hausdorff-Young inequality and some basic examples but with little success. Any help or advice would be helpful.

Answer 1

Finally, I have found an answer...

Using Plancherels identity on function $f(a,b,c)$, in group $ Z_3 $ we have: $$ \left( \dfrac{1}{3}\left( |a+b+c|^3+|a+\omega b +\omega^2c|^3+|a+\omega^2b+\omega c|^3\right) \right)^{\dfrac{1}{3}}\geq \left( \dfrac{1}{3}\left( |a+b+c|^2+|a+\omega b +\omega^2c|^2+|a+\omega^2b +\omega c|^2\right) \right)^{\dfrac{1}{2}} = \left(|a|^2+|b|^2+|c|^2\right)^{\dfrac{1}{2}}\geq \left( |a|^3+|b|^3+|c|^3\right)^{\dfrac{1}{3}} $$

Now it is clear that the problem holds.