How do I prove this?

How do I prove this?

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wolframalpha.com/input/?i=(a d + b c + b d)^2 - 4 a b c d > 0
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analytically

What does that mean?

It's false

How do you know?

where are the proofs?

Move 4abcd to the other side of the inequality, take sqare root of both sides. This should immediately look like a familiar inequality, with minor complications.

Why don't you just do the proof? What you said makes little sense.

let them all equal 1 lol

It's false, consider

a = -2b, -c/12 < d < 0

divide by (b d)^2 set A=a/b, C = c/d, get
(A-C)^2 + 2 A + 2 C, and it's obvious

>a = -2b

They're all positive.

That's what you're supposed to prove.

[math](1*1+1*1+1*1)^2 - 4*1*1*1*1 > 0 \\ (3)^2 - 4 > 0 \\ 9-4>0 \\ 5>0[/math]

No, you're trying to prove the left expression given the right.

sorry, should be
(A-C)^2 + 2 A + 2 C + 1 > 0
where A = a/b, C=c/d. Still obvious.

The original expression is just
((a/b-c/d)^2 + 2 a/b + 2 c/d + 1)*(b d)^2 >0

I'm not doing your homework, you fucking mongoloid. Do exactly what I said. It's a simple application of AM-GM.

Move 4abcd to the right hand side, expand the term in parentheses, then do some algebra and you should be fine.

But your statement is meaningless. What numbers am I allowed use? It isn't true if I can use negatives.

you can replace a,b,c,d>0 with the weaker condition
a/b + c/d > 0

I think you've done it my friend. Thanks.

Get a really big computer and try all combinations of numbers smaller than 10^200, easy proof. This is how the goldbach conjecture was proven.

wolframalpha.com/input/?i=(a d + b c + b d)^2 - 4 a b c d > 0

print this out and hand it to your teacher

(ad + bc)^2 - 4abcd > 0 is true when a,b,c,d > 0

(ad + bc + bd)^2 - 4abcd > 0

root ((ad + bc + bd)^2) >root( 4abcd)

(ad + bc + bd)^2 >2root(abcd)

Onli if a,b,c,d>0

ad + bc + bd >2root(abcd)

Only if a,b,c,d>0

You retard lmao