Question

Prove if x>0, y>0 and x 13 years ago

Answer 1

Consider variables x and y, where y is n larger than x.

y = x+n

Where n is any positive real number. x and y are real numbers greater than 1. This satisfies your conditions of x<y and x>0, y>0.

Therefore:

\[y^2 = (x+n)^2\]
expanding the right side:
\[y^2 = x^2 + nx + n^2\]

Therefore, for the assertion that x^2 < y^2 is true, then the following must also be true:
\[x^2 < x^2 + nx + n^2\]

The range of nx and n^2 for any positive, real n and any x>1 is always positive, therefore it is true.

Answer 2

y^2 - x^2 = (x+y)(y-x)

now (x+y)(y-x) > 0 always,
since x>0 and y>0
also y>x

which means
y^2 - x^2 > 0
or
y^2 > x^2 ..

Answer 3

Same as @shubhamsrg
Given: y>x
y-x>0
(y+x)(y-x)>(y+x)*0 [ since x>0 ,y>0 (x+y)>0]
(y^2-x^2)>0
y^2>x^2

Answer 4

[ since x>0 ,y>0 (x+y)>0]* the inequality symbol doesnot changes

Answer 5

thanks guys!