Today, we are going to derive some useful mathematical inequalities, starting with the fact that:

${\displaystyle \left(x-y\right)^{2n}\ge0}$ for ${\displaystyle n \in \N}$ (1).

Inequalities where the LHS is of the form ${\displaystyle x^{2n}+y^{2n}}$:

Such inequalities can be easily obtained by expanding the LHS of (1) and placing all terms not of the form ${\displaystyle x^{2n}}$ or ${\displaystyle y^{2n}}$ on the RHS. Doing this generates a sequence of identities:

${\displaystyle x^{2n}+y^{2n}\ge x^{2n}-\left(x-y\right)^{2n}+y^{2n}}$. (2)

Some examples of identities derived from this are:

${\displaystyle x^{2}+y^{2}\ge 2xy}$ (2 for ${\displaystyle n=1}$)

${\displaystyle x^{4}+y^{4}\ge 4x^{3}y-6x^{2}y^{2}+4xy^{3}}$ (2 for ${\displaystyle n=2}$)

Identities of that form can be stacked in two ways, as outlined in the below headings:

Product:

One way to stack identities is by multiplying two or more terms of the form ${\displaystyle x^{2n}+y^{2n}}$ together, and then placing all terms not of the form ${\displaystyle x^{2n}}$ or ${\displaystyle y^{2n}}$ on the RHS. This leads to what I call an inequality chain. The length of the inequality chain is dependent on ${\displaystyle n}$ such that the length is at least ${\displaystyle p\left(n\right)+1}$, where ${\displaystyle p\left(n\right)}$ is the partition function. As an example, consider the stacked inequality for when ${\displaystyle n=2}$.

${\displaystyle x^{4}+y^{4}\ge4x^{3}y-6x^{2}y^{2}+4xy^{3}\ge2x^{3}y+2xy^{3}-2x^{2}y^{2}\ge2x^{2}y^{2}}$ (3)

Ratio:

Another way to stack identities is to divide one identity by another. Examples include:

${\displaystyle \frac{x^{4}+y^{4}}{x^{2}+y^{2}}\le2x^{2}-3xy+2y^{2}}$ (4)

Applications:

If ${\displaystyle x \left( t \right)}$ and ${\displaystyle y \left( t \right)}$ have all numbers as their range, then this can be used to prove inequalities without calculus. One example we will show exploits two of the tricks used in the blog post.

Example:

Prove that ${\displaystyle \sin\left(x\right)^{4}+\cos\left(x\right)^{4}\ge\frac{1}{2}}$ without using calculus methods.

Note: Thie proof defines ${\displaystyle f=\sin\left(x\right)}$ and ${\displaystyle g=\cos\left(x\right)}$.

Proof:

The inequality we strive to prove becomes ${\displaystyle f^{4}+g^{4}\ge\frac{1}{2}}$.

Note that we can divide by ${\displaystyle f^{2}+g^{2}}$ without affecting the RHS side since ${\displaystyle f^{2}+g^{2}=1}$.

The equation then becomes: ${\displaystyle \frac{f^{4}+g^{4}}{f^{2}+g^{2}}\ge\frac{1}{2}}$.

We use (4), which will turn the LHS to ${\displaystyle 2f^{2}-3fg+2g^{2}\ge\frac{1}{2}}$.

Further simplification gives ${\displaystyle 2-3fg\ge\frac{1}{2}}$.

Solving for ${\displaystyle 2fg}$ gives us: ${\displaystyle 2fg\le1}$.

From the double-angle identity, ${\displaystyle 2fg=\sin\left(2x\right)}$, so this is saying ${\displaystyle \sin\left(2x\right)\le1}$, which is true since the amplitude of the sinusoid is 1. This is true.

If still not convinced, remember that ${\displaystyle 2fg\le f^{2}+g^{2}=1}$.

This concludes the proof.

For the Record:

This problem can be solved using single-variable calculus methods:

${\displaystyle f\left(x\right)=\sin\left(x\right)^{4}+\cos\left(x\right)^{4}}$

${\displaystyle f'\left(x\right)=-4\left(\cos\left(x\right)\right)^{3}\sin\left(x\right)+4\cos\left(x\right)\left(\sin\left(x\right)\right)^{3}}$

${\displaystyle 0=-4\left(\cos\left(x\right)\right)^{3}\sin\left(x\right)+4\cos\left(x\right)\left(\sin\left(x\right)\right)^{3}}$

${\displaystyle x=\frac{n\pi}{4}}$

${\displaystyle \left(\sin\left(\frac{\pi}{4}\right)\right)^{4}+\left(\cos\left(\frac{\pi}{4}\right)\right)^{4}=\frac{1}{2}}$

It can also be solved using Lagrangian multipliers, using the function to be minimized ${\displaystyle x^{4}+y^{4}}$ with the constraint ${\displaystyle x^2 + y^2 = 1}$:

${\displaystyle \lambda<4x^{3},4y^{3}>=<2x,2y>}$

${\displaystyle <4\lambda x^{3},4\lambda y^{3}>=<2x,2y>}$

${\displaystyle 4x^{3}-2x=0,4y^{3}-2y=0}$

${\displaystyle x=\left[-\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\right]}$

${\displaystyle y=\left[-\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\right]}$

${\displaystyle \left(-\frac{1}{\sqrt{2}}\right)^{4}+\left(-\frac{1}{\sqrt{2}}\right)^{4}=\frac{1}{2}}$

Concluding Notes:

As shown above, those inequalities are powerful and simple to generate. They can also be used to solve inequality problems in mathematical contests or can be used to do calculus homework without the calculus. I will see you in the next blog post concerning a new way to approximate functions relying on Taylor polynomials. Enjoy!

-chikorita the recreational mathematician