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)
