Question

设a,b,c>0,且(a+b-c)(1/a+1/b-1/c)=4.求证：(a^4+b^4+c^4)(1/a^4+1/b^4+1/c^4)>=2304.

Answer 1

证明：注意到(a+b-c)(1/a+1/b-1/c)=3+a/b-a/c+b/a-b/c-c/a-c/b=4
∴a/b-a/c+b/a-b/c-c/a-c/b=1，即a/b+b/a=b/c+c/b+c/a+a/c+1
令a/b+b/a=x，b/c+c/b=y，c/a+a/c=z，那么x=y+z+1，且yz=(b/c+c/b)(c/a+a/c)=b/a+a/b+ab/c²+c²/(ab)≥x+2
又yz≤(y+z)²/4=(x-1)²/4
∴x+2≤(x-1)²/4，即(x+1)(x-7)≥0，又x>0，∴x≥7，∴y+z=x-1≥6
(a^4+b^4+c^4)(1/a^4+1/b^4+1/c^4)=3+(a/b)^4+(b/a)^4+(b/c)^4+(c/b)^4+(a/c)^4+(c/a)^4
注意到(a/b)^4+(b/a)^4=[((a/b)²+(b/a)²]²-2=(x²-2)²-2
∴(a^4+b^4+c^4)(1/a^4+1/b^4+1/c^4)=(x²-2)²+(y²-2)²+(z²-2)²-3≥(x²-2)²+(y²+z²-4)²/2-3≥(x²-2)²+[(y+z)²/2-4]²/2-3≥(7²-2)²+(6²/2-4)²/2-3=2304