can anybody help me doing this lil math question ?

Question

if a and b are both positive
demonstrate that : Va+b < Va+Vb
V = square root
thnxxxxxx

Answer 1

Do you mean √(a+b) < √a + √b?

In that case, just notice that a + b < a + b + 2√(ab) = (√a + √b)^2. Then take the square root of both sides (since everythings positive).

Answer 2

Since:

Va
therefore Va + Vb
Then it must mean is always Va+Vb

Answer 3

hey smartpantys, I'll take a stab at this:

For this comparative equation to work, the variable B must be greater than 1. example, a=16, b=1. So, we have:
Va + b < Va + Vb.
V16+1< V16=V1,
4+1<4+1
5<5, wrong.

But when variable b is greater than 1, the comparison equation works very well. do i need to demonstrate, or will you just trust me.... It's late....OK

Answer 4

√(a+b) < √a + √b
then square both sides
[ √(a+b) ]^2 < [ √a + √b ]^2
a + b < a + 2√ab + b
a + b < a + b + 2√ab

Answer 5

square both sides you get for √[a+b] ^ 2 = a+b for √a + √b you get a+b+2√[ab]; as long as 2√[ab] > 0, the inequality is true.

Answer 6

I assume you mean
sqrt(a+b) < sqrt(a) + sqrt(b)
so
square both sides and expand the right
a+b < a + 2sqrt(a*b) + b
subtract (a+b)
0

Answer 7

if you mean you need to demonstrate that sqrt(a+b) sqrt(a+b)sqrt(a+b)^2<(sqrt(a)+sqrt(b))^2
<=>a+b 0<2sqrt(a*b) <=> 0

Answer 8

i cant help