Inequalities

Theorem 1:

I shall introduce a notation here: $\forall$ . It means “for all”. Now,

$x^2 \geq 0, \forall x \in R$

This inequality is extremely useful as a lot of complex expressions are solved through this inequality.

*Interesting fact: There exists an imaginary number that satisfy the condition $x^2 = -1$ . (Let the number be $x$ .) For those who are eager to know, this ‘number’ is often used in solving the roots of a quadratic equation in the form of $ax^2+bx+c=0 (a,b,c \in R, a \not= 0)$ when $b^2-4ac < 0$ . Theorem 2: When $a > b$ , then $a \pm c > b \pm c$ .

Theorem 3:

When $a > b$ , then $ac > bc$ for $c > 0$ .

When $a > b$ , then $ac < bc$ for $c < 0$ . Theorem 4: When $a > b$ , then $\displaystyle{\frac{a}{c} > \frac{b}{c}}$ for $c > 0$ .

When $a > b$ , then $\displaystyle{\frac{a}{c} < \frac{b}{c}}$ for $c < 0$ . Now a simple application of the basic inequalities we just learnt. Problem 1 Prove that $x^4 - x^2y^2 + y^4 \geq 0$ for all real $x$ and $y$ , with equality holding if and only if $x = y = 0$ . (Solutions at the end of the post) The theorem above is mostly needed to prove. Now, we shall learn an also very important inequality, namely the Quadratic Inequality. This is often needed to solve for unknowns. Theorem 5 When $(ax - m)(bx - n) > 0$ ( $a,b,m,n$ are real constants, $a,b \neq 0$ ), the solutions for $x$ are $\displaystyle{x > \frac{m}{a}}$ and $\displaystyle{x < \frac{n}{b}}$ , with the condition of $\displaystyle{\frac{m}{a} > \frac{n}{b}}$ .

Theorem 6

When $(ax - m)(bx - n) < 0$ ( $a,b,m,n$ are real constants, $a,b \neq 0$ ), the solutions for $x$ are $\displaystyle{\frac{n}{b} < x < \frac{m}{a}}$ , with the condition of $\displaystyle{\frac{m}{a} > \frac{n}{b}}$ .

This is particularly useful when solving quadratic involving inequalities.

For example, a simple application would be:

$x^2-5x+3\geq-3$

After simplifying, one will get

$(x-2)(x-3)\geq0$

By applying the formula, the solution for this inequality will be

$x \leq 2$ and $x \geq 3$ .

There is also something called the “Mean Inequality”, however it is not widely used in lower or even upper secondary works, mainly due to the fact the it is only used in proving equations, and we are often told to “solve” equations. If you are interested in it, you can go search more about it.

Solution 1

By completing the squares, we get:

$x^4-x^2y^2+y^4 = (x^2-y^2)^2+(xy)^2$

Clearly then, the LHS of the equation is non-negative, and is $0$ iff $x^2=y^2$ and $xy=0$ .

Therefore, the only possibility for the LHS of the equation to be 0 is when $x=y=0$ . (You can derive this from solving the simultaneous equations $x^2=y^2$ and $xy=0$ )

Cheers,

etzhky =)