demonstrate that :
√(a+b) < √a + √b
thnxxxxxxx
2006-12-01 20:42:14 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
It isnt true for every a and b.
For instance, if a = 4 and b = 0 the inequality becomes:
√(4+0) < √4 + √0
2<2
which isnt true.
[If either a≤0 or b≤0 then it isnt true]
If a and b are both positive then it is true, since in that case,
ab > 0
√(ab) > 0
2√(ab) > 0
a + b + 2√(ab) > a + b
[√a + √b]^2 > a + b
√a + √b > √(a + b)
2006-12-01 21:16:23 · answer #1 · answered by Scott R 6 · 3⤊ 1⤋
Aaahhaa! You changed the question, now, which I previously answered to perfect satisfaction. And now you bait and switch.
You are lower than a used car salesman, lower than a Democratic Representative in Congress! Slink back to your previous question, and give best answer to me for it, or I will call you "Smartpantys" from now on!!!
2006-12-02 06:06:04 · answer #2 · answered by Anonymous · 0⤊ 0⤋
For example:
â(a+b) < âa + âb
â(9+16) < â9 + â16
â25 < â9 + â16
5 < 3 + 4
5 < 7
.......
2006-12-02 05:01:06 · answer #3 · answered by aeiou 7 · 0⤊ 4⤋
â(a+b) < âa + âb
then square both sides
[ â(a+b) ]^2 < [ âa + âb ]^2
a + b < a + 2âab + b
a + b < a + b + 2âab
2006-12-02 04:53:01 · answer #4 · answered by starstrucktv 2 · 2⤊ 3⤋
From this i only understand
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2006-12-02 08:18:35 · answer #5 · answered by Mikhil M 2 · 0⤊ 0⤋