if a,b,c, are in h.p. prove that a^2+c^2> 2b^2 ?
answer= 1/a,1/b, and 1/c are in h.p.
=1/b-1/a=1/c-1/b
=1/b+1/b=1/c+1/a
=b+b/b^2=a+c/ac
=2/b^2=a+c/ac
=b^2(a+c)=2ac
=(a+c)ac=2b^2
=a^2+b^2 >2b^2
google is your friend. This should help.
http://www.askiitians.com/forums/Algebra/22/42109/progressions.htm
To prove that a^2 + c^2 > 2b^2 when a, b, and c are in harmonic progression (H.P.), we can follow the following steps:
1. Assume that a, b, and c are in harmonic progression, which means that 1/a, 1/b, and 1/c are in arithmetic progression (A.P.).
2. Start by writing the given harmonic progression relationship in terms of reciprocals: 1/b - 1/a = 1/c - 1/b.
3. Simplify the equation: (a - b) / (ab) = (c - b) / (bc).
4. Cross-multiply to eliminate fractions: (a - b) * (bc) = (c - b) * (ab).
5. Expand both sides: abc - b^2c = acb - b^2a.
6. Rearrange the terms: abc - acb = b^2c - b^2a.
7. Factor out common terms: ac(b - c) = b^2(c - a).
8. Divide both sides by (b - c) and cancel common terms: ac = b^2.
9. Substitute this result back into the original equation: (a + c) * ac = 2b^2.
10. Simplify further: a^2 + ac + ac + c^2 = 2b^2.
11. Combine like terms: a^2 + 2ac + c^2 = 2b^2.
12. Notice that 2ac is positive since all variables are positive.
13. Thus, we can conclude that a^2 + c^2 > 2b^2, as desired.
Therefore, it has been proved that a^2 + c^2 > 2b^2 when a, b, and c are in harmonic progression (H.P.).
To prove that a^2 + c^2 > 2b^2 when a, b, and c are in harmonic progression (H.P.), we need to show that (a + c)ac = 2b^2.
Let's start by defining the harmonic progression (H.P.). In an H.P., the reciprocals of the terms are in arithmetic progression (A.P.). Therefore, we can express a, b, and c as 1/a, 1/b, and 1/c, respectively, such that 1/a, 1/b, and 1/c are in an A.P.
Using this information, we can rewrite the given inequality as:
(a + c)ac > 2b^2 (dividing both sides by b^2)
Now, let's use the fact that 1/a, 1/b, and 1/c are in an A.P. to rewrite the above expression:
[(1/a) + (1/c)] * (1/a) * (1/c) > 2(1/b)^2
Simplifying the expression, we get:
[(1/a) + (1/c)] * (1/a) * (1/c) > 2/(b^2)
Now, cross-multiplying, we have:
(1/a + 1/c) * (1/a) * (1/c) - 2/(b^2) > 0
Expanding the expression, we get:
(1/a^2 + 1/ac + 1/ac + 1/c^2) - 2/(b^2) > 0
Combining the terms with common denominator, we have:
(1/a^2 + 2/ac + 1/c^2) - 2/(b^2) > 0
Now, let's simplify further:
(a^2 + 2ac + c^2) - 2/(b^2) > 0
Multiplying through by a^2b^2c^2 (since a, b, and c are non-zero), we have:
(a^2 + 2ac + c^2)(a^2b^2c^2) - 2 > 0
Expanding the expression, we get:
a^4b^2c^2 + 2a^3b^2c^3 + a^2b^2c^4 - 2 > 0
Now, let's factorize (a^4b^2c^2 + 2a^3b^2c^3 + a^2b^2c^4) as a perfect square:
(a^2bc + ac^2)^2 - 2 > 0
We can see that (a^2bc + ac^2)^2 is always positive, so for the inequality to hold true, we need:
(a^2bc + ac^2)^2 > 2
Since (a^2bc + ac^2)^2 is always greater than or equal to 0, the inequality holds true.
Therefore, we can conclude that a^2 + c^2 > 2b^2 when a, b, and c are in harmonic progression.