中四程度pure math 5題(大陸)－rll33xb99t的部落格

標題:

中四程度pure math 5題(大陸)

發問:

証明下列各恆等式: 1: 2(a^3+b^3+c^3)>bc(b+c)+ca(c+a)+ab(a+b) 2: 1/a +1/b +1/c>1/(sqrt bc) + 1/(sqrt ca) +1/(sqrt ab) 3: (a+b+c)(a^2+b^2+c^2)>9abc 4: 2/(b+c) +2/(c+a) +2/(a+b)>9/(a+b+c) 5: (bc)^2+(ca)^2+(ab)^2>abc(a+b+c) 拜託了~!

最佳解答:

1. WLOG, let a>b>c By Rearrangement Inequality a3 + b3 > a2b + b2a = ab(a+b) --- (i) Similarly, b3 + c3 > bc(b+c) --- (ii) a3 + c3 > ac(a+c) --- (iii) (i)+(ii)+(iii), the result follows 2. By AM-GM Inequality 1/a + 1/b = (a+b)/ab > 2(sqrt ab) / ab = 2/(sqrt ab) --- (i) Similarly, 1/b + 1/c > 2/(sqrt bc) --- (ii) 1/a + 1/c > 2/(sqrt ac) --- (iii) (i)+(ii)+(iii) 2(1/a + 1/b + 1/c) > 2/(sqrt ab) + 2/(sqrt bc) + 2/(sqrt ac) The result follows 3. By AM-GM Inequality (a+b+c)(a2+b2+c2) > [3 (abc)^(1/3)] x [3 (a2b2a2)^(1/3)] = 9abc 4. By AM-HM Inequality 2/(a+b) + 2/(b+c) + 2/(c+a) > 3 x {3 / [ (a+b)/2 + (b+c)/2 + (a+c)/2 ]} = 3 x [3 / (a+b+c)] = 9/(a+b+c) 5. By AM-GM Inequality (ab)2+ (bc)2 > 2sqrt[(ab)2+(bc)2] = 2ab2c --- (i) Similarly, (bc)2+ (ca)2 > 2bc2a --- (ii) (ca)2+ (ab)2 > 2ca2b --- (iii) (i)+(ii)+(iii) (ab)2+ (bc)2 + (ca)2 > abc(a+b+c) Hope the above information helps =) By 小儒