i need help proving this inequality

pepsigirl

That {x/[x^2+y^2]^1/2} <1

Reply 1

rbnphlp

The bottom bit is always positive whatever the case because they are both squares added together .

Reply 2

AsakuraMinamiFan

Work backwards by multiplying both sides by sqrt(x^2 + y^2), squaring, etc but remember when reversing your argument that taking the square root has two solutions.

Reply 3

Ry4n11

pepsigirl

That {x/[x^2+y^2]^1/2} <1

If y=0, the inequality would read "1 < 1".

Which is not true, and therefore this cannot be proven?

Reply 4

pepsigirl

Ry4n11

If y=0, the inequality would read "1 < 1".

Which is not true, and therefore this cannot be proven?

yes... the question doesn't specify for which values, so i just took it to be all... obviously not

Reply 5

AsakuraMinamiFan

It holds for all y not equal to 0.

Reply 6

qgujxj39

I think you wanted a

\leq

sign rather than a

<

, this makes the inequality true when

y = 0

.

When

x \leq 0

, the statement is clearly true, because the LHS is negative/zero (the denominator is always positive), so you only need to consider the case for positive x. This means you can square both sides and still preserve the inequality.

We can also consider this geometrically: if we have the points A(0,0), B(x,0) and C(x,y) then clearly AC > AB (if