output_description: If an N-sided polygon satisfying the condition can be drawn, print `Yes`;
  otherwise, print `No`.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      L_1 L_2 ... L_N
attempt: # -*- coding: utf-8 -*-
  
  #############
  # Libraries #
  #############
  
  import sys
  
  input = sys.stdin.readline
  
  import math
  import bisect
  from collections import deque
  from fractions import gcd
  from functools import lru_cache
  
  
  #############
  # Constants #
  #############
  
  MOD = 10**9 + 7
  INF = float("inf")
  
  #############
  # Functions #
  #############
  
  
  ######INPUT######
  def inputI():
      return int(input().strip())
  
  
  def inputS():
      return input().strip()
  
  
  def inputIL():
      return list(map(int, input().split()))
  
  
  def inputSL():
      return list(map(str, input().split()))
  
  
  def inputILs(n):
      return list(int(input()) for _ in range(n))
  
  
  def inputSLs(n):
      return list(input().strip() for _ in range(n))
  
  
  def inputILL(n):
      return [list(map(int, input().split())) for _ in range(n)]
  
  
  def inputSLL(n):
      return [list(map(str, input().split())) for _ in range(n)]
  
  
  ######OUTPUT######
  def Yes():
      print("Yes")
  
  
  def No():
      print("No")
  
  
  #####Inverse#####
  def inv(n):
      return pow(n, MOD - 2, MOD)
  
  
  ######Combination######
  kaijo_memo = []
  
  
  def kaijo(n):
      if len(kaijo_memo) > n:
          return kaijo_memo[n]
      if len(kaijo_memo) == 0:
          kaijo_memo.append(1)
      while len(kaijo_memo) <= n:
          kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
      return kaijo_memo[n]
  
  
  gyaku_kaijo_memo = []
  
  
  def gyaku_kaijo(n):
      if len(gyaku_kaijo_memo) > n:
          return gyaku_kaijo_memo[n]
      if len(gyaku_kaijo_memo) == 0:
          gyaku_kaijo_memo.append(1)
      while len(gyaku_kaijo_memo) <= n:
          gyaku_kaijo_memo.append(
              gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo), MOD - 2, MOD) % MOD
          )
      return gyaku_kaijo_memo[n]
  
  
  def nCr(n, r):
      if n == r:
          return 1
      if n < r or r < 0:
          return 0
      ret = 1
      ret = ret * kaijo(n) % MOD
      ret = ret * gyaku_kaijo(r) % MOD
      ret = ret * gyaku_kaijo(n - r) % MOD
      return ret
  
  
  ######Factorization######
  def factorization(n):
      arr = []
      temp = n
      for i in range(2, int(-(-(n**0.5) // 1)) + 1):
          if temp % i == 0:
              cnt = 0
              while temp % i == 0:
                  cnt += 1
                  temp //= i
              arr.append([i, cnt])
      if temp != 1:
          arr.append([temp, 1])
  
      if arr == []:
          arr.append([n, 1])
      return arr
  
  
  #####LCM#####
  def lcm(a, b):
      return a * b // gcd(a, b)
  
  
  #####BitCount#####
  def count_bit(n):
      count = 0
      while n:
          n &= n - 1
          count += 1
      return count
  
  
  #############
  # Main Code #
  #############
  
  N = inputI()
  l = inputIL()
  l.sort()
  if sum(l[:-1]) >= sum[l]:
      No()
  else:
      Yes()
  
problem_description: Statement
  
  Determine if an N-sided polygon (not necessarily convex) with sides of length
  L_1, L_2, ..., L_N can be drawn in a two-dimensional plane.
  
  You can use the following theorem:
  
  **Theorem** : an N-sided polygon satisfying the condition can be drawn if and
  only if the longest side is strictly shorter than the sum of the lengths of
  the other N-1 sides.
samples: [{"input": "4\n    3 8 5 1", "output": "Yes\n    \n\nSince 8 < 9 = 3 + 5 + 1, it follows from the theorem that such a polygon can\nbe drawn on a plane.\n\n* * *"}, {"input": "4\n    3 8 4 1", "output": "No\n    \n\nSince 8 \\geq 8 = 3 + 4 + 1, it follows from the theorem that such a polygon\ncannot be drawn on a plane.\n\n* * *"}, {"input": "10\n    1 8 10 5 8 12 34 100 11 3", "output": "No"}]