d****t
发帖数: 26 | | k**y
发帖数: 320 | 2 the key thing is let x=b+c, y=a+c, z=a+b and want to prove:
x/y+y/z+z/x+y/x+z/y+x/z>=6. you can either try
(x/y+y/z+z/x)+(y/x+z/y+x/z)>=6 or try
(x/y+y/x)+(z/x+x/z)+(y/z+z/y)>=6
every correct proof so far did the first way i guess. | b****l
发帖数: 132 | 3 Another proof:
a^2 - ab + b^2 >= ab --> a^3 + b^3 >= ab(a + b)
a^2 - ac + c^2 >= ac --> a^3 + c^3 >= ac(a + c)
b^2 - bc + c^2 >= bc --> b^3 + c^3 >= bc(b + c)
add the 3 inequations up:
2a^3 + 2b^3 + 2c^3 > ab(a+b) + ac(a+c) + bc(b+c) -->
2a^3 + 2b^3 + 2c^3 + 2abc > ab(a+b) + ac(a+c) + bc(b+c) + 2abc -->
a^3 + b^3 + c^3 + abc 1 |
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