#include<algorithm>
#include<complex>
#include<ctype.h>
#include<iomanip>
#include<iostream>
#include<fstream>
#include<map>
#include<math.h>
#include<numeric>
#include<queue>
#include<set>
#include<stack>
#include<stdio.h>
#include<string>
#include<string>
#include<vector>

using namespace std;
typedef long long ll;

#define FOR(i,a,b) for(ll i=(a);i<(b);++i)
#define ALL(v) (v).begin(), (v).end()
#define p(s) cout<<(s)<<endl
#define p2(s, t) cout << (s) << " " << (t) << endl
#define br() p("")
#define pn(s) cout << (#s) << " " << (s) << endl
#define p_yes() p("Yes")
#define p_no() p("No")

const ll mod = 1e9 + 7;
const ll inf = 1e18;

bool can_make(ll r, ll g, ll num){
    if(r>g){
        swap(r, g);
    }

    ll b = 0;
    // rを十分作る
    if(r<num){
        ll diff = num - r;
        g -= 2 * diff;
    }
    // rが十分にある
    else{
        ll diff = r - num;
        b = diff / 2;
    }

    ll diff = num - b;
    if(diff>0){
        // bを十分作る
        g -= 2 * diff;
    }

    if(g>=num){
        return true;
    }else{
        return false;
    }
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    // input
    vector<ll> A(3);
    FOR(i, 0, 3){
        cin >> A[i];
    }
    sort(ALL(A));

    ll num = A[0];
    
    ll r = A[1] - A[0];
    ll g = A[2] - A[0];

    // 追加で何個作れるかを二分探索
    ll left = 0;
    ll right = 10000000;
    while(left+1!=right){
        ll center = (left+right)/2;
        if(can_make(r, g, center)){
            left = center;
        }else{
            right = center;
        }
    }
    
    p(num + left);
    
    return 0;
}