Inequalities (1 Viewer)

[azureus88]

Use the fact that [maths]f(x)=(ax-b)^2+(cx-d)^2\geq 0[/maths] to prove [maths]|ab+cd|\leq \sqrt{a^2+c^2}\sqrt{b^2+d^2}[/maths] (a,b,c,d real)

It's obvious if you just use [maths]2abcd\leq a^2d^2+b^2c^2[/maths] to prove [maths]\sqrt{(ab+cd)^2}\leq \sqrt{(a^2+c^2)(b^2+d^2)}[/maths] but how would you do the question using f(x) ?

 

[untouchablecuz]

here you go

edit:

it shoodnt be "there are no intercepts with the x axis"

but rather "there may be either 1 or 0 intercepts, hence discriminant <= 0"

 

[Timothy.Siu]

well done

 

[shaon0]

here you go

Thats an elegant solution. I would give u rep but i can't because i have to "pass it around."

 

[azureus88]

thanks

 

[khorne]

I believe this is the closest you will ever get to the maths version of street cred.

 

[kwabon]

here you go

edit:

it shoodnt be "there are no intercepts with the x axis"

but rather "there may be either 1 or 0 intercepts, hence discriminant <= 0"

awesome work, would have never thought of that.

 

[untouchablecuz]

lol, thanks everyone :]

 

[Affinity]

Use the fact that [maths]f(x)=(ax-b)^2+(cx-d)^2\geq 0[/maths] to prove [maths]|ab+cd|\leq \sqrt{a^2+c^2}\sqrt{b^2+d^2}[/maths] (a,b,c,d real)

It's obvious if you just use [maths]2abcd\leq a^2d^2+b^2c^2[/maths] to prove [maths]\sqrt{(ab+cd)^2}\leq \sqrt{(a^2+c^2)(b^2+d^2)}[/maths] but how would you do the question using f(x) ?

Classic Proof for the Cauchy Schwarz inequality. You can also substitute:
[maths] x = \dfrac{ab+cd}{a^2 + c^2} [/maths] into f(x)

Now try this

Use the fact that [maths]f(x)=|ax-b|^2+|cx-d|^2\geq 0[/maths] to prove [maths]|a\bar{b}+c\bar{d}|\leq \sqrt{(|a|^2+|c|^2)(|b|^2+|d|^2)}[/maths] where a,b,c,d are complex numbers