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[Akshara Patil]

 

[jimmysmith560]

Would the following working help?

Given that the AM/GM inequality is true for n = k.

We have to prove that the inequality will be true for n = 2k.

That means we have to show: x1 + x2 + ... +x2k2k ≥ x1x2 ... x2k2k.

Since the inequality is true for n = k values. So if we take the first k values then we get:

x1 + x2 + ... + xkk ≥ x1x2 ... xkk .............(1).

Similarly for the last k values then we get: xk + 1 + xk + 2 + ... + x2kk ≥ xk + 1xk + 2 ... x2kk ..................(2).

Then we add (1) and (2):

x1 + x2 + ... + xkk + xk + 1 + xk + 2 + ... + x2kk ≥ x1x2 ... xkk + xk + 1xk + 2 ... x2kkx1 + x2 + ... + xk + xk + 1 + xk + 2 + ... + x2kk ≥ x1x2 ... xkk + xk + 1xk + 2 ... x2kkx1 + x2 + ... + x2kk ≥ x1x2 ... xkk + xk + 1xk + 2 ... x2kk ...............(3)

We know that:

a - b2 ≥ 0 or, a2 + b2 - 2ab ≥ 0 or, a2 + b2 ≥ 2ab ..............(4)

Let, a = x1x2 ... xk2k and b = xk + 1xk + 2 ... x2k2k

Using (4) we get:

x1x2 ... xk2k2 + xk + 1xk + 2 ... x2k2k2 ≥ 2x1x2 ... xk2kxk + 1xk + 2 ... x2k2k or, x1x2 ... xkk + xk + 1xk + 2 ... x2kk ≥ 2x1x2 ... xk2kxk + 1xk + 2 ... x2k2k

Note: p2k2 = p12k2 = p1k = pk or, x1x2 ... xkk + xk + 1xk + 2 ... x2kk ≥ 2x1x2 ... xkxk + 1xk + 2 ... x2k2k

Because mr · nr = mnr or, x1x2 ... xkk + xk + 1xk + 2 ... x2kk ≥ 2x1x2 ... x2k2k .................(5)

Combining (3) and (5) we get:

x1 + x2 + ... + x2kk ≥ x1x2 ... xkk + xk + 1xk + 2 ... x2kk ≥ 2x1x2 ... x2k2k or, x1 + x2 + ... + x2kk ≥ 2x1x2 ... x2k2k Removed middle partor, x1 + x2 + ... + x2k2k ≥ x1x2 ... x2k2k Divide both sides by 2

Hence, we have shown that, if the AM/GM inequality is true for n = k, then it will be true for n = 2k. [Proved (i)]