# Minkowski's and Holder's inequality confusion

Something doesn't completely makes sense to me in Minkowski's inequality and I'm trying to understand it.

All throughout let $a_i, b_i \geq 0$, and $p > 1$, $\frac{1}{p} + \frac{1}{q} = 1$.

Holder's inequality states that:

$\sum_i a_ib_i \leq (\sum_i a_i ^p)^\frac{1}{p} (\sum_i b_i ^q)^\frac{1}{q}$

And the equality happens exactly when $a_i ^p = b_i^q$.

I'm trying to figure out when Minkowski's inequality turns into an equality.

Minkowski states $(\sum_i (a_i + b_i)^p)^\frac{1}{p} \leq (\sum_i a_i ^p)^\frac{1}{p} + (\sum_i b_i ^p)^\frac{1}{p}$

One of the stages of the proof is stating that $\sum_i (a_i + b_i)^p = \sum_ia_i(a_i+b_i)^{p-1} + \sum_i b_i (a_i + b_i)^{p-1}$ and now using Holder's inequality on these two summations which will lead to an inequality.

But when is that inequality an equality? Since we used Holder, it should be the same criteria but applied twice:

$a_i ^p = ((a_i+b_i)^{p-1})^{q}$ from the first sum

$b_i ^p = ((a_i+b_i)^{p-1})^{q}$ from the second

after realizing that $q = \frac{p}{p-1}$ this simplifies to $a_i ^p = b_i^ p = (a_i+b_i)^p$

Seems to me as if the only solution to these equations is $a_i = b_i = 0$. However, since this is just the triangle inequality, the criteria should be that $a = kb$ for some constant $k$, but I'm not seeing it based on this proof.

Can someone shed light on the matter?

You can check that your condition does not cover the case where $a_i>0$ but $b_i =0$, $\forall i$.

The point is, the condition that Holder inequality becomes equality is

$$\alpha a_i^p = \beta b_i^q,$$

for some numbers $\alpha,\beta$, while at least one of them should be non-zero. Now by this condition, the equality conditions of Minkowski inequality are $\exists k_1,k_2,h_1,h_2$ s.t. $k_1^2 + k_2^2 >0$, $h_1^2+h_2^2>0$, and

$k_1a^p_i=k_2((a_i+b_i)^{p−1})^q = k_2(a_i+b_i)^{p}$ from the first sum

$h_1b^p_i=h_2((a_i+b_i)^{p−1})^q = h_2(a_i+b_i)^{p}$ from the second

In the non-trival case, $k_1,k_2,h_1,h_2>0$, and the two conditions imply that $a_i = kb_i$.