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Note sure about the "best" solution, but here's how I'd do it in C++:

#include <string>
#include <iostream>

using namespace std;

bool is_palindrome(const string &inputString)
{
    for(int i = 0, j = inputString.size() - 1; i < j; i++, j--)
        if (inputString[i] != inputString[j]) return false;
    return true;
}

int main(int argc, char *argv[])
{
    if (argc != 2) cout << "Usage: " << argv[0] << " <string>" << endl;
    else cout << "The string '" << argv[1] << "' is "
                << (is_palindrome(argv[1]) ? "" : "not ")
                << "a palindrome." << endl;
    return 0;
}

But since you mentioned close-to-the-metal, I endeavored to do it in C instead:

#include <stdio.h>
#include <string.h>

int is_palindrome(const char *inputString)
{
    int i, j;

    if (inputString)
        for(i = 0, j = strlen(inputString) - 1; i < j; i++, j--)
            if (inputString[i] != inputString[j]) return 0;
    return 1;
}

int main(int argc, char *argv[])
{
    if (argc != 2) printf("Usage: %s <string>\n", argv[0]);
    else printf("The string '%s' is %sa palindrome.\n",
                argv[1], (is_palindrome(argv[1]) ? "" : "not "));
    return 0;
}

Note that in both cases, the loop conditional in is_palindrome also ensures that the empty string qualifies as a palindrome, without improperly attempting to access a character at index -1.