More Proofs Involving Inequalities

[turophile]

I think I have figured out how to prove: If a > b > 0, then a^2 > b^2. It goes something like this:

a > b > 0 ==> a^2 > ab
a > b > 0 ==> ab > b^2
Since a^2 > ab and ab > b^2, then a^2 > b^2.

I'm having more trouble with these proofs:

1. If a > b > 0, then sqrt(a) > sqrt(b).

2. If x > y, then x^3 > y^3.

Any hints?

 

[soroban]

Hello, turophile!

\(\displaystyle 2.\;\text{ If }x > y,\,\text{ then }\,x^3 \:>\: y^3\)

\(\displaystyle \text{Suppose }x\text{ and }y\text{ have opposite signs.}\)
\(\displaystyle \text{Then }x\text{ is positive and }y\text{ is negative.}\)
\(\displaystyle \text{Hence, }x^3\text{ is positive and }y^3\text{ is negative.}\)

\(\displaystyle \text{Therefore: }\:x^3 \:>\:y^3\)


\(\displaystyle \text{Suppose }x\text{ and }y\text{ have the same sign.}\)
\(\displaystyle \text{Then }xy\text{ is positive.}\)

\(\displaystyle \text{We have: }\:x\:>\:y\)
. . \(\displaystyle \text{Multiply by }x^2\!:\;\; x^3 \:>\:x^2y\) .[1]

\(\displaystyle \text{We have: }\:x\:>\:y\)
. . \(\displaystyle \text{Multiply by }xy\!:\;\; x^2y \:>\:xy^2\) .[2]

\(\displaystyle \text{We have: }\:x\:>\:y\)
. . \(\displaystyle \text{Multiply by }y^2\!:\;\; xy^2 \:>\:y^3\) .[3]

From [1], [2] and [3], we have: . \(\displaystyle x^3 \:>\:x^2y \:>\:xy^2\:>\:y^3\)

\(\displaystyle \text{Therefore: }\:x^3 \:>\:y^3\)

 

[turophile]

Thanks soroban. Should I also prove for the cases where x = 0 and y = 0? Something like this?

Let x = 0
x = 0 ? x^3 = 0
x > y and x = 0 ? y < 0 ? y^3 < 0
? x^3 > y^3

Let y = 0
y = 0 ? y^3 = 0
x > y ? x > 0 ? x^3 > 0
? x^3 > y^3

Therefore if x > y, then x^3 > y^3 for all real numbers x and y.