A twist on a classic inequality: Can You Solve It?
The only difference between algebra and kindergarten math is using symbols instead of pure numbers.
Both are instances of computation; mathematicians strive for symbolic beauty, but that beauty can also be found in plain-old numbers. Sometimes, when nothing works out, we go back to what we know from kindergarten.
Today’s problem demonstrates this statement:
It makes use of the C-S inequality, given by:
The Solution:
There’s something very useful (and difficult to notice without a keen eye): The number of terms is the same as the sum of terms!
Now, with such an important condition, you should make full use of it. The question is, how?
Actively searching for hidden inequalities is key. Here’s something important:
The C-S inequality needs a product. No product exists in the given expressions, at least no obvious products.
Here, when it seems extremely difficult to find existing products, you make the products.
This is an artificial product, created from scratch; it closely resembles the right-hand side of the C-S inequality. The best thing is, multiplying by 1 won’t affect anything at all!
You’re close, but not quite. The inequality has been reduced to squares, and the same steps can easily be repeated:
The desired result follows immediately.
That’s a normal solution explaining all the “correct thinking”: getting to the answer in the shortest number of steps possible.
Often times, the amount of incorrect work that ends up getting discarded is several times the number of correct steps; these mistakes typically fail to make it to the reader.
Now, for the curious reader, here was my full thought process:
I started by thinking that this was an easy problem. My first thought to construct squares was overly synthetic: by simply squaring both sides.
It was ugly, but I wasn’t willing to give up. After some attempts at using C-S, I soon realized how this was an obvious mistake, something which was going to lead me nowhere. The construction was too strong, and it led me to avoid squares for the rest of the problem.
Again, inspired by previous problems, I attempted a fractional construction. This was closer to the desired outcome, giving me
I was immediately concerned by this: the problem didn’t say a ≠ 0. Still, I didn’t want to give up this possibility too easily.
I attempted the bounding through direct application of C-S:
If we wanted this to work, we would need to bound the second sum on the right hand side. That easily failed, so I started from scratch, observing possible constructions.
On the third attempt, I found the correct construction, and everything proceeded smoothly from there.
It’s that simple. Have fun problem solving!
