Show that from a group of seven people whose (integer) ages add up to 332 one can select three people with the total age at least 142. I need help with this problem, and I was thinking in this way:
$$ x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} = 332 $$
and I need to find three of these which sum is at least 142.
But I don't know what next. 
Any help would be appreciated
 A: The average of the age sum over all possible triples is three times the average age per person (since each person occurs the same number of times). Since the average age per person is above $142\,/\,3$, the average age sum per triple is above $142$, so at least one triple must have age sum at least $142$.
A: Suppose that any $3$ chosen people have at most a total age of $141$. Then, $x_1 + x_2 + x_3 \le 141$ and so is $x_4 + x_5 + x_6$. This indicates that $x_7 \ge 50$. 
Repeating the same process exchanging the roles of $x_i$'s yields $x_i \ge 50$ for any $i$. Which leads to a contradiction.
A: You can form $\binom{7}{3}$ triplets, i.e. 35 triplets. Every men participates in $\binom{6}{2}$ triplets, i.e 15. So the sum of all 35 triplets is $15 \times 332 = 4980$, which means the average is $\frac{4980}{35} = 142.427...$ so there has to be at least one triplet with total age of at least 142. Actually since all years are integers this proves that there has to be a triplet with total sum of at least 143.
A: Assume $x_1\le x_2\le\ldots\le x_7$.  Here's a proof that $x_5+x_6+x_7\ge144$ (which answers an observation made by TonyK in comments).  
Suppose $x_5+x_6+x_7\le143$.  Then $x_1+x_2+x_3+x_4\ge332-143=189$, which implies $x_4\ge189/4=47.25$.  Since the ages are assumed to be integers, we have $48\le x_4\le x_5\le x_6\le x_7$, which contradicts the supposition.
A: I find the complexity of the methods shown here to be strange. There is a much simpler solution. $332$ years divided among $7$ people works out to be a little more than $47$ years per person. Specifically, because $7 \cdot 47 = 329$, there are three extra years. So the most even distribution of ages is:
$$47,47,47,47,48,48,48.$$
Obviously, there is a group of three whose ages add up to at least $142$ (in fact, it is $144$). Further note that any time you take a year from someone, you must give it to someone else to keep the sum constant. Taking a year from an older person just redistributes the years and taking a year from a younger person does not help in trying to get a lower sum.
To elaborate on this, you can do one of four redistributions at the beginning: from a $47$ to a $47$, from a $47$ to a $48$, from a $48$ to a $47$, and from a $48$ to a $48$. After these, the outcomes (when sorted again) will be:
$$46,47,47,48,48,48,48$$
$$46,47,47,47,48,48,49$$
$$47,47,47,47,48,48,48$$
$$47,47,47,47,47,48,49$$
As you can see, in every case, the three greatest ages have a sum that is at least $3 \cdot 48 = 144$. Now note that the next time you take a year from someone and give it to someone else, if the first person is older, then you will get one of the five configurations seen already. The only way to get a configuration not seen before is to take a year from a younger person, and this will never lower the sum of the ages of the three oldest people.
To look at it another way, if every person was $47$ years old, then every triple of persons has an age sum of $141$. The instant you give anyone an extra year, there is a triple with an age sum of $142$. Adding two more extra years doesn't help.
A: Proof by contradiction :
Suppose that it is impossible to find 3 such numbers. Then for every triplet $ x_{i}, x_{j}, x_{k}  \in   \{x_{1}, ... , x_{7}\}, x_{i} + x_{j} + x_{k} < 142  $.
Thus we have $ \sum_{i=1}^{7} x_{i} < 142 +  \sum_{i=4}^{7} x_{i} $ $(*)$
But we also have $ \sum_{k=4,k \ne j }^{7} x_{k} < 142, \forall j \in \{4,5,6,7\}  $.
Summing these for all $j \in \{4,5,6,7\}  $ we get $ 3 * (\sum_{i=4}^{7} x_{i}) < 4*142 $ and then $   (\sum_{i=4}^{7} x_{i}) < \frac{4}{3}*142 = 189.333 .. < 190 $ .
Replacing this in $(*)$ we get $ \sum_{i=1}^{7} x_{i} < 190 + 142 = 332 $, which is a contradiction.
A: Proof by contradiction. Given $\sum_{i=1}^{7} x_i = 332$, assume that we cannot pick three people whose ages sum to at least 142. This means that for all $i$, $j$, $k \in [1, 7]$ with $i \ne j$, $j \ne k$, and $i \ne k$, $x_i + x_j + x_k < 142$. With these two assumptions,


*

*$x_1 + x_2 + x_3 < 142$ and $x_4 + x_5 + x_6 < 142$ implies $x_7 > 48$.

*$x_1 + x_2 + x_3 < 142$ and $x_5 + x_6 + x_7 < 142$ implies $x_4 > 48$.

*$x_2 + x_3 + x_4 < 142$ and $x_5 + x_6 + x_7 < 142$ implies $x_1 > 48$.
But $x_1 + x_4 + x_7 > 3 \times 48 = 144 > 142$, which contradicts the assumption that for all $i$, $j$, $k$ with $i \ne j$, $j \ne k$, and $i \ne k$, $x_i + x_j + x_k < 142$. There there exists at least one combination of $i, j, k \in [1, 7]$ with $i \ne j$, $j \ne k$, and $i \ne k$, $x_i + x_j + x_k > 142$.
A: Given seven integers with sum $332$, label them such that $x_1 \geq x_2 \geq x_3 \geq x_4 \geq x_5 \geq x_6 \geq x_7$. Proof by decomposition into two cases: $x_3 \geq 48$ and $x_3 \lt 48$.
If $x_3 \geq 48$ then $x_1 + x_2 + x_3 \geq 48 + 48 + 48 = 144$ (by choice of labels). Therefore in this case $x_1 + x_2 + x_3 \gt 142$.
If $x_3 \lt 48$, or rather $x_3 \le 47$, then $x_4 + x_5 + x_6 + x_7 \le 47 + 47 + 47 + 47 = 188$ (by choice of labels). So we have $x_1 + x_2 + x_3 \ge 332 - 188 = 144$, therefore $x_1 + x_2 + x_3 \gt 142$.
A: Consider all of the three-element subsets of the seven numbers. The average sum of each subset must be 3/7 · 332 &approx; 142.3. If no subset sums to 142 or more, then every subset must be below average, which is impossible.
A: Proof by contradiction.
If you cannot select 3 elements in the set whose some is bigger or equal 142, for every triplet $ x_{i}, x_{j}, x_{k}  \in   \{x_{1}, ... , x_{7}\}, x_{i} + x_{j} + x_{k} < 142  $.
This implies:
$ x_{1}, x_{2}, x_{3} < 142 $,
$ x_{1}, x_{2}, x_{4} < 142 $,
..
$ x_{1}, x_{2}, x_{7} < 142 $,
$ x_{2}, x_{3}, x_{4} < 142 $,
.. and so on up to
$ x_{5}, x_{6}, x_{7} < 142 $.
If you sum up all the left and right side of the 35 inequalities (combining triplets in $ \{x_{1}, ... , x_{7}\} $ without repetitions) you will obtain a new inequality
$ 15 \sum_{i=1}^{7} x_{i} < 35 \times 142 $
that can be rewritten as
$ \sum_{i=1}^{7} x_{i} < \frac{35 \times 142}{15} $
i.e.
$ \sum_{i=1}^{7} x_{i} < 331.33$
but this is not possible as the text of the problem states that
$ \sum_{i=1}^{7} x_{i} = 332 $.
So this means that our initial assumption:
$ x_{i}, x_{j}, x_{k}  \in   \{x_{1}, ... , x_{7}\}, x_{i} + x_{j} + x_{k} < 142  $
is wrong and therefore it should exists at least a triplet whose sum is $ >= 142 $ q.e.d.
A: You need 7 inequations.
Suppose you can't select three people with the total age at least 142.
Then:
x1+x2+x3<142 (eq. 1)
   x2+x3+x4<142 (eq. 2)
      ...
           x5+x6+x7<142 (eq. 5)
x1+             x6+x7 <142 (eq. 6)
x1+x2+            +x7 <142 (eq. 7)
Then
3x1+3x2+....+3x7<7x142
=> x1+x2+...+x7<7x142/3<332 =><=
A: The average integer age of all $7$ is $47$. $3 \times 47 = 141$. However, because $7 \times 47 = 329$, it follows that one person must have an age of $50$ in order for the total to be $332$. Take $2$ average people and the old guy and you have $144$.
