inequations questions (1 Viewer)

[bos1234]

Are these methods correct/any quicker ones. Thanks bye

Given that x< y< 0, show that xy > y^2
y^2-xy<0
y(y-x)<0
y<x<0

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Suppose x>y>0, show that x^2 > y^2
x^2-y^x>0
(x+y)(x-y)>0
x-y>0
x>y>0

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Prove that x^2 + xy + y^2 > 0 for any non-zero value of x and y

x^2 + xy + y^2 > 0
(x+y)^2 -xy > 0
x^2 + 2xy + y^2 -xy > 0
x^2+xy+y^2>0

thanks bye

 

[vulgarfraction]

Not really...

The first one just wasn't right in the first place. Counterexample (-1, 1)

The second is fundamental right, just in the wrong direction. Correct proof:

x>y>0
=> x+y>0, x-y>0
=>(x+y)(x-y)>0
=>x^2-y^y>0 as required

The third one... just look at what you did again. Closely.

(Correct proof: multiply everything by 2, express as sum of 3 squares)

Hope that helps.

 

[bos1234]

sry i typed the 1st question wrong.. ill type it properly later

ok thanks

 

[bos1234]

Not really...

The first one just wasn't right in the first place. Counterexample (-1, 1)

The second is fundamental right, just in the wrong direction. Correct proof:

x>y>0
=> x+y>0, x-y>0
=>(x+y)(x-y)>0
=>x^2-y^y>0 as required

The third one... just look at what you did again. Closely.

(Correct proof: multiply everything by 2, express as sum of 3 squares)

Hope that helps.

how did u get the step in bold.. and i re wrote the 1st question...

 

[elseany]

if both x and y are positive then their sum must be positive.
if x is greater than y, then the difference between x and y (i.e. x-y) must also be positive.

but remember its x-y which is positive. y-x would be negative.

 

[ronaldinho]

for the last qn.. can u do it like this??????? going around in cricles i think?

x^2 + xy + y^2 > 0 .......A
(x+y)^2-xy > 0
(x+y)^2 > xy

A....(x+y)^2 -xy > 0
since (x+y)^2 > xy
then (x+y)^2 -xy is greater than 0

PLEASE POST UP OTHER SOLNS..

 

[SoulSearcher]

I always liked doing the last question this way, not sure it's right though:

x³-y³ = (x-y)(x²+xy+y²)
When x > y,
x³-y³ is positive, x-y is positve, and as such x²+xy+y² is positive to maintain the sign of the product, thus it is greater than 0.
When x < y,
x³-y³ is negative, x-y is negative and as such x²+xy+y² is positive to maintain the sign, and similar to above, greater than 0.
When x = y,
x³-y³ = 0, x-y=0 but x²+xy+y² = 3x² > 0 since a square of a number is always positive.

Therefore x²+xy+y² > 0 for x and y where x and y are non-zero values.

Not entirely sure it's correct, but looks it.

 

[ronaldinho]

yeah i think that method is correct. But its very complicated

how about the method above?Is that right or wrong?

x^2 + xy + y^2 > 0 .......A
(x+y)^2-xy > 0
(x+y)^2 > xy B

Factorising A. (x+y)^2 -xy > 0
since (x+y)^2 > xy from B
then (x+y)^2 -xy is greater than 0




Here is another way posted by Mark576

we know that x^2 and y^2 are greater than 0, and that x^2 +2xy + y^2 > 0, so:
(x^2 +y^2)/2 +((x+y)^2)/2 > 0, simplifying this, it becomes:
x^2 + xy + y^2 > 0

 

[vulgarfraction]

x^2 + xy + y^2 = 1/2(x^2 + y^2 + (x+y)^2) >= 0

first one:

y <0, (x-y) <0
=> y(x-y)<0
=> xy > y^2 as required