My proof: If a > b, then, successively,
(a - b) > 0, so (a - b) / 2 > 0, .........(A)
so, adding (a + b) / 2 to each side:
a > (a + b) / 2. QED (i)
Similarly, adding 'b' to each side of equation (A),
(a + b) / 2 > b. QED (ii)
Thus, a > (a + b) / 2 > b. QED (i) and (ii) combined.
Live long and prosper.
2007-01-05 07:06:41
·
answer #1
·
answered by Dr Spock 6
·
0⤊
0⤋
this is not achievable to factorize RHS to have a ingredient that has LHS. because you may want to finally favor to practice that RHS = LHS + X the position X ? 0. inspite of the indisputable fact that you'll practice it by yet differently it truly is to thoroughly improve RHS and evaluate it with the higher LHS to practice a similar element. it truly is RHS = LHS + X the position X ? 0. inspite of the indisputable fact that this is not classy and intensely tedious. So for lazy ppl, like me, we'd want to favor to practice it by merely searching and evaluating it. considering a b c are +ve actual numbers (mutually with 0), enable c ? b ? a ? 0 be real, then as u can see, the smallest fee a,b or c can carry is 0. consequently because the guy lower than proved it, (0.5 - 0^2 +3)^3 ? (0 + 0 +0 )^3 => 27 is more suitable than 0 (real truth) a^5 ? a^2 IFF a takes on a ? a million, for this reason we ought to analyze 2 circumstances for a ? a million and a million > a ? 0 (fractions) for a ? a million, a^5 ? a^2 consequently, a^5-a^2 might want to consistently be ? 0 and minimum fee of RHS is 27 for a million > a ? 0 a^2 ? a^5 and a^5-a^2 < 3, infact it would want to consistently be < -a million. for this reason, a^5 - a^2 + 3 might want to consistently be +ve (it would want to border of mind a huge decision bt for this qn we are able to not favor to noe wad tat huge form is; ploting the graph and the minimum of x +ve might want to be it) considering, the above might want to be conscious for both the b and the c factors of the RHS, RHS is continually +ve even as a, b, c are +ve actual numbers. and now for the LHS, even as a,b,c are ? a million, RHS ? LHS as RHS has the further 3 besides because the better skill. even as a million > c ? b ? a ? 0 (they're all fractions): even as a techniques 0 from a million, (a^5 - a^2 + 3) might want to be U-formed graph from 0 to at least a million, the position the minimum is about 2.6743 even as a about 0.7368. it is the same for b and c. consequently, RHS is continually ? LHS
2016-12-01 21:15:03
·
answer #2
·
answered by Anonymous
·
0⤊
0⤋
The easiest way may be to break it into two parts; first proving a>(a+b)/2, then proving that (a+b)/2>b. Starting with what you know, a>b, add a to the inequality, giving you a+a >a+b, or 2a>a+b, divide by 2, a > (a+b)/2. Proving the other half works similarly. (a>b, a+b > b+b, a+b > 2b, (a+b)/2 > b).
2007-01-05 07:18:08
·
answer #3
·
answered by Walter B 2
·
0⤊
0⤋
a>b => a > (a+b)/2 > b
First Part: a > b
a+a > a + b
2a > a + b
a > (a + b)/2
Second part: a > b
a + b > b + b
a + b > 2b
(a + b)/2 > b
Third part : a > (a + b)/2 and (a + b)/2 > b so
a > (a + b)/2 > b
qed
2007-01-05 08:48:24
·
answer #4
·
answered by frank 7
·
0⤊
0⤋
I think I thought of a way.
You're given that a > b.
Therefore a - b > 0.
If you add 2b to both sides:
a - b + 2b > 2b
a + b > 2b
Divide by 2:
(a + b)/2 > b -- that's half of it.
Again starting with:
a - b > 0
If you multiply through by -1, you get:
b - a < 0
Now add 2a to both sides:
b - a + 2a < 2a
b + a < 2a
Divide both sides by 2:
(a + b)/2 < a
That's the other half.
Therefore a > (a + b)/2 > b.
QED.
2007-01-05 06:56:06
·
answer #5
·
answered by Jim Burnell 6
·
0⤊
1⤋
a>b then a>(a+b)/2>b
2>0 then 2>(2+0)/2>0 , so 2>1>0 with the universe being Z+
2007-01-05 06:58:20
·
answer #6
·
answered by Anonymous
·
0⤊
1⤋
let a-b=x>0
(a+b)/2=(2b+x)/2=b+x/2. Since x>0, x/2>o and
(a+b)/2>b
also (a+b)/2=(2a-x)/2=a-x/2 since x/2>0, -x/2<0 therefore
a>1-x/2
and a>(a+b)/2>b
2007-01-05 07:15:40
·
answer #7
·
answered by yupchagee 7
·
0⤊
0⤋
it is true. lets say that a is 4 and b is 2. a>b is true now, 2+4 is 6 then divide it by 2 and ya get 3, 4>3>2 a>(a+b)/2>b 4>(4+2)/2>2 it is true.
2007-01-05 06:47:21
·
answer #8
·
answered by Anonymous
·
0⤊
1⤋
It is common sense really.
a>(a+b)/2>b is merely stating that a is greater than the mean value between a and b. Obviously, if a is greater than b, then it will also be greater than a value in between them.
2007-01-05 06:45:58
·
answer #9
·
answered by Dendryte88 4
·
0⤊
1⤋
prove it?
a is greater than b because every sign says a is greater every time
2007-01-05 06:46:22
·
answer #10
·
answered by Anonymous
·
0⤊
3⤋