This is another solution by Yatir from Maccabim-Reut High
School, Israel.
First let's prove that the sum of a positive number and its
reciprocal is at least 2. It is clearly true for all x that
(x-1)2 ³ 0, hence x2-2x+1 ³ 0 and it follows that x2 + 1 ³ 2x.
Since x > 0 there is no harm in dividing by x proving that for all x > 0
Another challenge is to use the hints given by the two
illustrations in the question and to give alternative proofs that
the sum of a positive number and its reciprocal is greater than
or equal to 2.
Yatir uses this inequality when he sums k fractions and their
reciprocals in the following proof. Can you use a similar method
to give a shorter proof of the result without resorting to
mathematical induction? You will need to expand the expression
given in (1), collect pairs of terms, decide how many pairs there
are and use the inequality for each of the pairs of terms.
We have to prove that:
|
(x1 + x2 + ¼+ xn)( |
1
x1
|
+ |
1
x2
|
+ ¼+ |
1
xn
|
) ³ n2. (1) |
|
Let's carry on by induction, first for n=1 we have
so this part is proved.
Let's assume that it is true for a positive integer n=k:
|
(x1 + x2 + ¼+ xk)( |
1
x1
|
+ |
1
x2
|
+ ¼+ |
1
xk
|
) ³ k2. |
|
Now let's prove it for n=k+1. We have
|
(x1 + x2 + ¼+ xk + xk+1)( |
1
x1
|
+ |
1
x2
|
+ ¼+ |
1
xk
|
+ |
1
xk+1
|
) |
|
|
= (x1 + x2 + ¼+ xk)( |
1
x1
|
+ |
1
x2
|
+ ¼+ |
1
xk
|
) |
| |
|
+ xk+1( |
1
x1
|
+ |
1
x2
|
+ ¼+ |
1
xk
|
) |
| |
|
+ |
1
xk+1
|
(x1 + x2 + ¼+ xk) |
| |
| |
| ³ k2 + |
k
å
i=1
|
( |
xk+1
xi
|
+ |
xi
xk+1
|
)+ 1. |
|
|
We have used the induction assumption that the result is true for
n=k in the last line. By opening this up we see that we have the
sum of k positive numbers and their reciprocals. Since each one
of them is at least 2 we replace each of them by 2 so the
inequality becomes:
| (x1 + x2 + ¼+ xk + xk+1)( |
1
x1
|
+ |
1
x2
|
+ ¼+ |
1
xk
|
+ |
1
xk+1
|
) ³ k2 + 2k + 1 = (k+1)2 |
|
|
This proves the inequality for n=k+1 and so, by the axiom of
induction, the result is true for all positive integers.