вариант 2
-6 + (-8) = - 14
-6-8 = -14
-6+8 = 2
-6-(-8) = 2
-(-6)-(-8) = 14
-(-6)+8 = 14
-(-6)+(-8) = -2
-(-6)-8 = -2
6-8 = -2
6+8 = 14
6-(-8) = 14
6+(-8) = -2
-8-6 = -14
-8-(-6) = -2
-8+6 = -2
-8+(-6) = -14
8-6 = 2
8-(-6) = 14
8+6 = 14
8+(-6) = 2
-(-8)-6 = 2
-(-8)+6 = 14
-(-8)+(-6) = 2
-(-8)-(-6) = 14
вариант 1
-4+(-9) = -13
-4-9 = -13
-4+9 = 5
-4-(-9) = 5
-(-4)-(-9) = 13
-(-4)+9 = 13
-(-4)+(-9) = 5
-(-4)-9 = -5
9-4 = 5
9-(-4) = 13
9+4 = 13
9+(-4) = 5
4-9 = 5
4+9 = 13
4-(-9) = 13
4+(-9) = -5
-9-4 = -13
-9-(-4) = -5
-9+4 = -5
-9+(-4) = -13
-(-9)-4 = 5
-(-9)+4 = 13
-(-9)+(-4) = 5
-(-9)-(-4) = 13
Должно быть a>0, b>0, c>0, d>0, а так же 1/a+1/b+4/c+16/d≥64/(a+b+c+d)
Пошаговое объяснение:
1/a+2/b+8/c+16/d≥64/(a+b+c+d)⇔(a+b+c+d)(1/a+2/b+8/c+16/d)≥64
Докажем последнее неравенство используя неравенство Коши-Буняковского
(a₁²+a₂²+a₃²+...+aₓ²)(b₁²+b₂²+b₃²+...+bₓ²)≥(a₁b₁+a₂b₂+a₃b₃+...+aₓbₓ)²
(a+b+c+d)(1/a+2/b+8/c+16/d)=
=((√a)²+(√b)²+(√c)²+(√d)²)((1/√a)²+(√(2/b))²+(√(8/c))²+(√(16/d))²)≥
(√a·1/√a+√b·√2/b+√c√(8/c)+(√d√(16/d))²=(√1+√2+√8+√16)=(5+3√2)²>
>(5+3√1,96)²=(5+3·1,4)²=9,2²=84,64⇒(a+b+c+d)(1/a+2/b+8/c+16/d)>64
Доказано, что (a+b+c+d)(1/a+2/b+8/c+16/d)>64⇒
⇒1/a+2/b+8/c+16/d>64/(a+b+c+d)
3 9/17+2/17 = 3 11/17,
1 1/7+5/7 = 1 6/7,
3 1/5+ 5 2/5 = 8 3/5,
3 1/2+ 1 1/2 = 5,
4 3/5+1 2/5 = 6,
5 3/5+1 3/5 = 7 1/5,
3 2/7+2 6/7 = 6 1/7,
16 3/8+7 1/8 = 23 4/8 = 23 1/2,
17 2/9+9 4/9 = 26 6/9 = 26 2/3,
2 1/4+1 1/4 = 3 2/4 = 3 1/2,
7 1/3+2 1/3 = 9 2/3,
2 1/2+1/6 = 2 3/6 + 1/6 = 2 4/6 = 2 2/3,
3 7/12+1/6 = 3 7/12 + 2/12 = 3 9/12 = 3 3/4,
3 3/4+1/5 = 3 15/20 + 4/20 = 3 19/20,
7 9/20+7/30 = 7 27/60 + 14/60 = 7 41/60,
9 1/2+3 1/8, = 9 4/8 + 3 1/8 = 12 5/8,
6 9/16+2 1/4 = 6 9/16+ 2 4/16 = 8 13/16,
9 2+1 1/4 =
13 1/5+4 2/7 = 13 7/35 + 4 10/35 = 17 17/35