Prove or disprove this inequality

[kakarukeys]

a^2 - b^2 - c^2 - d^2 > 0
A^2 - B^2 - C^2 - D^2 = 1
A > 1
a > 0
Prove or disprove Aa + Bb + Cc + Dd > 0

after 3 days of trying I give up
can anyone give a clue?

 

[kakarukeys]

what I have done:
I used (a + b)^2 = a^2 + b^2 + 2ab, but it leads to nothing.
I have tried Lagrange Multipliers, but the only extremum is a saddle point.

 

[kakarukeys]

trying to disprove it
I found no counter-example.

 

[TenaliRaman]

(a+A)^2 + (b+B)^2 + (c+C)^2 + (d+D)^2 > 0

-- AI

 

[kakarukeys]

I can't see.
then?

 

[Motifs]

Are b,c,d >0 and B,C,D>1 ?? if so, no need to prove anything since it is obvious.
If not, I think we can't say the given formula >0 or <0
You can try subsitute some values of b,c,d to guess a, then values for B,C,D to calculate A, you will see it.

Am I incorrect ?

 

[Wong]

Let me try to give some motivation about the inequaility.

Consider the equivalent problem in 2D. That is,
a^{2}-b^{2}&gt;0, a&gt;1
A^2-B^2=1, A&gt;1
Prove or disprove aA+bB&gt;0
In 2D, the region represented by the first equation is like a "quadrant" rotated by 45 degrees. (Try to plot it.) The set represented by the second equation is one component of a hyperbola included in the first region. Note that Aa+Bb is just the usual *dot product* between vector (a, b) and (A, B). Can you see why the inequality holds in the case?

Try to do the problem for 3D and then generalise it.

 

[Motifs]

Wow, my gal bit me .

 

[kakarukeys]

To Wong,

The angle between the vectors is always less than 90?
I can see it in 2D and 3D, but not in 4D, but how to write a proof?

 

[Wong]

hi kakarukeys.

To prove it, first note that the set represented by a^{2}-b^{2}-b^{2}-c^{2}=1, a&gt;1 is in fact a subset of A^2-B^2-C^2-D^2&gt;0, A&gt;0. Then what we need to prove becomes
Prove Aa+Bb+Cc+Dd&gt;0, where (A, B, C, D) and (a, b, c, d) both satisfies,
h^2-i^2-j^2-k^2&gt;0, h&gt;0
A^2&gt;B^2+C^2+D^2
A&gt;(B^2+C^2+D^2)^{\frac{1}{2}}
Similarly,
a&gt;(b^2+c^2+d^2)^{\frac{1}{2}}
Put those expression into Aa+Bb+Cc+Dd, does it remind you of some inequality?
This can be generalised to any dimensions.

Somtimes it can be quite difficult to think of a proof for inequalities, even though it is quite trivial geometrically.

 

[kakarukeys]

Intuitive guide, I have spotted the 'subset' clue.
but I still can't see the solution.

Aa &gt; \sqrt{B^2 + C^2 + D^2}\sqrt{b^2 + c^2 + d^2} \geq Bb + Cc + Dd
(Cauchy-Schwarz inequality)

And so Aa - Bb - Cc - Dd &gt; 0
that's not we want.

Note Bb + Cc + Dd can be negative.

 

[kakarukeys]

ok COOL
Aa &gt; \sqrt{B^2 + C^2 + D^2}\sqrt{b^2 + c^2 + d^2} = \sqrt{(-B)^2 + (-C)^2 + (-D)^2}\sqrt{b^2 + c^2 + d^2}<br /> \geq - Bb - Cc - Dd

 

[kakarukeys]

Thumbs up!

 

[TenaliRaman]

Interesting indeed!
I was lost with my initial thoughts !
Bravo!

-- AI