Question

If + =+/−
, where x is real, then prove that, 2 +2 = 1 and /=2/2−1
.

Answer 1

Answer:

$2xy ≤ x2 + y2$

Step-by-step explanation:

Proof. First we prove that if x is a real number, then x2 ≥ 0. The product

of two positive numbers is always positive, i.e., if x ≥ 0 and y ≥ 0, then

xy ≥ 0. In particular if x ≥ 0 then x2 = x · x ≥ 0. If x is negative, then −x

is positive, hence (−x)2 ≥ 0. But we can conduct the following computation

by the associativity and the commutativity of the product of real numbers:

$0 ≥ (−x)2 = (−x)(−x) = ((−1)x)((−1)x) = (((−1)x))(−1))x= (((−1)(x(−1)))x = (((−1)(−1))x)x = (1x)x = xx = x2$

The above change in bracketting can be done in many ways. At any rate,

this shows that the square of any real number is non-negaitive. Now if x and

y are real numbers, then so is the difference, x − y which is defined to be

x + (−y). Therefore we conclude that 0 ≤ (x + (−y))2 and compute:

$0 ≤ (x + (−y))2 = (x + (−y))(x + (−y)) = x(x + (−y)) + (−y)(x + (−y)) = x2 + x(−y) + (−y)x + (−y)2 = x2 + y2 + (−xy) + (−xy) = x2 + y2 + 2(−xy);$

adding 2xy to both sides,

$2xy = 0 + 2xy ≤ (x2 + y2 + 2(−xy)) + 2xy = (x2 + y2) + (2(−xy) + 2xy)= (x2 + y2) + 0 = x2 + y2$

Therefore, we conclude the inequality:

$2xy ≤ x2 + y2$

for every pair of real numbers x and y.