(click anywhere to toggle day/night mode)
BACKWe want to prove that
$$\sum_{k=1}^{n} \left | x_{k}y_{k}\right |\leqslant \left [ \sum_{k=1}^{n} \left | x_{k}\right | ^{p}\right ]^{1/p}\left [ \sum_{k=1}^{n} \left | y_{k}\right |^{q}\right ]^{1/q}$$
Where $$p,q$$ are real numbers and $$\frac{1}{p} + \frac{1}{q} = 1$$
First, starting from cases with the same power. Let's use 3. We want to prove that: $$\sqrt[3]{\sum a_{i}^{3}}\sqrt[3]{\sum b_{i}^{3}}\sqrt[3]{\sum c_{i}^{3}} \geq \sum a_{i}b_{i}c_{i}$$
Let's say (I know one would ask "why???" This is not my idea but one would see why after they see the whole proof) $$A = \sqrt[3]{\sum a_{i}^{3}}$$, and the same for B and C. Let's also say $$x_{i} = \frac{a_{i}}{A}$$ and the same for y and z.
Now this is very good for us because $$\sum x_{i}^3 = 1$$ That's because if you sum up all the x, the denominator and the numerator becomes $$A^3 = \sum a_{i}^{3}$$
We wanted to prove $$ABC \geq \sum a_{i}b_{i}c_{i}$$ With substitution, we know that now we want to prove that $$\sum x_{i}y_{i}z_{i} \leq 1$$
Using AM-GM on $$x_{i}^3, y_{i}^3, z_{i}^3$$ For each i, we have $$\frac{x_{i}^3+y_{i}^3+z_{i}^3}{3} \geq \sqrt[3]{x_{i}^3y_{i}^3z_{i}^3} = x_{i}y_{i}z_{i}$$
Now let's find the sum of the left hand side: $$\sum \left ( \frac{x_{i}^{3}+y_{i}^{3}+z_{i}^{3}}{3} \right )$$ We know that $$\sum x_{i}^3 = 1$$ and it's also the same for y and z. So the sum of them is 3 and it is divided by 3... the left hand side becomes 1.
We've proved that $$1\geq \sum x_{i}y_{i}z_{i}$$ And this is equivalent to the original inequality.
Now for a more common case, where there are k columns (imagine x, y, and z as columns,) but they still possess the same power. We want to prove: $$\sum_{i}^{}a_{1i}a_{2i}...a_{ki} \leq \prod_{j=1}^{k}\sqrt[k]{\sum_{i}^{}a_{ji}^{k}}$$
Say $$x_{ji} = \frac{a_{ji}}{\sqrt[k]{\sum_{i}^{}a_{ji}^{k}}}$$ so that $$\sum x_{ji}^k = 1$$
Notice that in the first part we used A,B, and C, which is equivalent to $$\sqrt[k]{\sum_{i}^{}a_{ji}^{k}}$$
For each i (meaning, the number with the same row in each column), we have $$\frac{x_{1i}^k + ... +x_{ki}^k}{k} \geq x_{1i}... x_{ki}$$ by AM-GM. I skipped the radical form because it is miserable to get the LaTex done.
As $$\sum x_{ji}^k = 1$$, we know the sum of the left hand side would be 1.
Therefore, $$\sum x_{1i}... x_{ki} \leq 1$$ Although the proof could be said as complete here, I thought it would be better that I unfold those x.
Because $$x_{ji} = \frac{a_{ji}}{\sqrt[k]{\sum_{i}^{}a_{ji}^{k}}}$$ this inequality means that $$\sum_{i=1} \prod_{j=1}^{k} a_{ji}\leq \prod_{j=1}^{k} \sqrt[k]{\sum_{i}^{}a_{ji}^k}$$ Which is what we wanted!
We've seen from the very start, the inequality is only true when $$\frac{1}{p} + \frac{1}{q} = 1$$
The key here is using weighted AM-GM: $$\frac{1}{q}x + \frac{1}{p}y \geq x^\frac{1}{q} y^\frac{1}{p}$$
Let's say $$A = \left(\sum a_{i}^q \right) ^ \frac{1}{q}, B = \left(\sum b_{i}^p \right) ^ \frac{1}{p}$$
And (we've done this so many times in this proof) $$x_{i} = \frac{a_{i}}{A}, y_{i} = \frac{b_{i}}{B}$$
Now we want to prove that $$\sum x_{i}y_{i} \leq 1$$
Here we want to use the weighted AM-GM. Substitute X and Y with $$X = x_{i}^q, Y = y_{i}^p$$ This is because only in this way the other side of the AM-GM is the product of x and y.
Then $$\frac{x_{i}^q}{q} + \frac{y_{i}^p}{p} \geq x_{i}y_{i}$$
$$\sum x_{i}y{i} \leq \frac{1}{q}\sum x_{i}^q + \frac{1}{p}\sum y_{i}^p $$
Because we know that $$\sum x_{i}^q = 1, \sum y_{i}^p = 1$$
$$\sum x_{i}y_{i} \leq \frac{1}{q} + \frac{1}{p} $$
Because the inequality is satisfied when $$\sum x_{i}y_{i} \leq 1$$, therefore$$\frac{1}{q} + \frac{1}{p} = 1$$
If you use the weighted AM-GM for more columns, you will be able to find the inequality satisfied even when dealing with an arbitrary number of columns and unequal exponents, as long as the reciprocals of the exponents add up to 1.
Some might say the prerequisite of the Hölder's inequality is so restricted and wonder what insight does it bring. It plays a vital role in Minkowski's inequality's proof (at least in this version) and although Minkowski's inequality also seems to be an irelevant fun puzzle of algebra, it is actually not. I should tell you the meaning of it after you read the proof.
It looks like this: $$ \left [ \sum_{k=1}^{\infty } \left | x_{k}+y_{k}\right |^p\right ]^\frac{1}{p} \leq \left [ \sum_{k=1}^{\infty }\left | x_{k}\right |^p \right ]^\frac{1}{p}+\left [ \sum_{k=1}^{\infty }\left | y_{k}\right |^p \right ]^\frac{1}{p}$$
We don't like dealing with a bundle of symbols. This $$ \left [ \sum_{k=1}^{\infty } \left | x_{k}+y_{k}\right |^p\right ]$$ is massive and we might just call it S.
Then we are trying to prove $$S^\frac{1}{p} \leq \left [ \sum_{k=1}^{\infty }\left | x_{k}\right |^p \right ]^\frac{1}{p}+\left [ \sum_{k=1}^{\infty }\left | y_{k}\right |^p \right ]^\frac{1}{p}$$
Looking at S, we see that it is equivalent to $$\sum_{k=1}^{\infty}(x_{k}+y_{k})(x_{k}+y_{k})^{p-1}$$
I would call this step the hardest step to think of in a proof. I would not be able to think of it if I had not read the proof before!
Multiply it out. We have $$\sum_{k=1}^{\infty}(x_{k})(x_{k}+y_{k})^{p-1} + \sum_{k=1}^{\infty}(y_{k})(x_{k}+y_{k})^{p-1}$$
Let's deal with the first part of S. That is, $$\sum_{k=1}^{\infty}(x_{k})(x_{k}+y_{k})^{p-1}$$ Now let's bring up the Hölder's inequality.
Plug in $$x_{k}, (x_{k}+y_{k})^{p-1}$$, and set a and b as the power. Let a = p because, why not? Then we have:
$$\sum_{k=1}^{\infty} x_{k}(x_{k}+y_{k})^{p-1} \leq (\sum_{k=1}^{\infty} x_{k}^{p})^\frac{1}{p}(\sum_{k=1}^{\infty}((x_{k}+y_{k})^{p-1})^{b})^\frac{1}{b}$$
Where $$\frac{1}{p} + \frac{1}{b} = 1$$
Let's examine the relation between p and b. $$\frac{1}{p} + \frac{1}{b} = 1, p+b = pb, p = b(p-1), b = \frac{p}{p-1}$$
This saves us. Look at the latter part of the inequality, $$(\sum_{k=1}^{\infty}((x_{k}+y_{k})^{p-1})^{b})^\frac{1}{b}$$
Plug in b. We get $$(\sum_{k=1}^{\infty}((x_{k}+y_{k})^{p})^\frac{p-1}{p}$$
Review our definition of S and notice that this is just $$S^\frac{p-1}{p}$$
So the first part of S is $$ \leq(\sum_{k=1}^{\infty}(x_{k}^p))^\frac{1}{p}S^\frac{1}{b}$$
With the same routine, the second part of S is $$\leq(\sum_{k=1}^{\infty}(y_{k}^p))^\frac{1}{p}S^\frac{1}{b}$$
Thank you Hölder!
We can now assemble the two parts of S. $$ S \leq ((\sum_{k=1}^{\infty}(y_{k}^p))^\frac{1}{p}+(\sum_{k=1}^{\infty}(x_{k}^p))^\frac{1}{p})S^\frac{1}{b}$$
Divided by $$S^\frac{1}{b}$$ on both sides, we see $$S^\frac{b-1}{b}\leq ((\sum_{k=1}^{\infty}(y_{k}^p))^\frac{1}{p}+(\sum_{k=1}^{\infty}(x_{k}^p))^\frac{1}{p})$$
Review the relation between p and b, this is equivalent to $$S^\frac{1}{p}\leq ((\sum_{k=1}^{\infty}(y_{k}^p))^\frac{1}{p}+(\sum_{k=1}^{\infty}(x_{k}^p))^\frac{1}{p})$$
And that is equivalent to the original inequality.
Does it remind you of something? The format of C <= A + B? Yes, the triangle inequality theorem! If you still don't see the meaning, try to plug in p=2:
$$ \left [ \sum_{k=1}^{\infty } \left | x_{k}+y_{k}\right |^2\right ]^\frac{1}{2} \leq \left [ \sum_{k=1}^{\infty }\left | x_{k}\right |^2 \right ]^\frac{1}{2}+\left [ \sum_{k=1}^{\infty }\left | y_{k}\right |^2 \right ]^\frac{1}{2}$$
The left hand side look like the formula for distance in Euclidean geometry. It turns out that the Minkowski inequality shows that for any p that is a real number greater than or equal to one, expression $$(\sum \left | x_{i} \right | ^{p})^\frac{1}{p}$$ can be used as a definition of distance as it follows the triangle inequality theorem. In plain English, that is, you always walk the least if you are walking straightly.