Java Dictionary Searcher

Splitting the input string every possible way is not going to finish in a reasonable amount of time if you want to support 20 or more characters. Here's a more efficient approach, comments inline:

public static void main(String[] args) throws IOException {
    // load the dictionary into a set for fast lookups
    Set<String> dictionary = new HashSet<String>();
    Scanner filescan = new Scanner(new File("dictionary.txt"));
    while (filescan.hasNext()) {
        dictionary.add(filescan.nextLine().toLowerCase());
    }

    // scan for input
    Scanner scan = new Scanner(System.in);
    System.out.print("Enter a string: ");
    String input = scan.next().toLowerCase();
    System.out.println();

    // place to store list of results, each result is a list of strings
    List<List<String>> results = new ArrayList<>();

    long time = System.currentTimeMillis();

    // start the search, pass empty stack to represent words found so far
    search(input, dictionary, new Stack<String>(), results);

    time = System.currentTimeMillis() - time;

    // list the results found
    for (List<String> result : results) {
        for (String word : result) {
            System.out.print(word + " ");
        }
        System.out.println("(" + result.size() + " words)");
    }
    System.out.println();
    System.out.println("Took " + time + "ms");
}

public static void search(String input, Set<String> dictionary,
        Stack<String> words, List<List<String>> results) {

    for (int i = 0; i < input.length(); i++) {
        // take the first i characters of the input and see if it is a word
        String substring = input.substring(0, i + 1);

        if (dictionary.contains(substring)) {
            // the beginning of the input matches a word, store on stack
            words.push(substring);

            if (i == input.length() - 1) {
                // there's no input left, copy the words stack to results
                results.add(new ArrayList<String>(words));
            } else {
                // there's more input left, search the remaining part
                search(input.substring(i + 1), dictionary, words, results);
            }

            // pop the matched word back off so we can move onto the next i
            words.pop();
        }
    }
}

Example output:

Enter a string: aman

a man (2 words)
am an (2 words)

Took 0ms

Here's a much longer input:

Enter a string: thequickbrownfoxjumpedoverthelazydog

the quick brown fox jump ed over the lazy dog (10 words)
the quick brown fox jump ed overt he lazy dog (10 words)
the quick brown fox jumped over the lazy dog (9 words)
the quick brown fox jumped overt he lazy dog (9 words)

Took 1ms

If my answer seems silly, it's because you're really close and I'm not sure where you're stuck.

The simplest way given your code above would be to simply add a counter for the number of words and compare that to the number of matched words

    int count = 0; int total = 0;
    Scanner phraseScan = new Scanner(segment);
    while (phraseScan.hasNext()) {
        total++
        String word = phraseScan.next();
        for (int i=0; i<dic.length; i++) {
            if (word.equalsIgnoreCase(dic[i])) count++;
        }
    }
    if(total==count) System.out.println(segment);

Implementing this as a hash-table might be better (it's faster, for sure), and it'd be really easy.

HashSet<String> dict = new HashSet<String>()
dict.add("foo")// add your data


int count = 0; int total = 0;
Scanner phraseScan = new Scanner(segment);
while (phraseScan.hasNext()) {
    total++
    String word = phraseScan.next();
    if(dict.contains(word)) count++;
}

There are other, better ways to do this. One is a trie (http://en.wikipedia.org/wiki/Trie) which is a bit slower for lookup but stores data more efficiently. If you have a large dictionary, you might not be able ot fit it in memory, so you could use a database or key-value store like a BDB (http://en.wikipedia.org/wiki/Berkeley_DB)