一道不等式竞赛题设a,b,c∈R+且a+b+c=3,求证(a²+3b²)/ab²(4-ab)+(b²+3c²)/bc²(4-bc)+(c²+3a²)/ca²(4-ac)≥4
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![一道不等式竞赛题设a,b,c∈R+且a+b+c=3,求证(a²+3b²)/ab²(4-ab)+(b²+3c²)/bc²(4-bc)+(c²+3a²)/ca²(4-ac)≥4](/uploads/image/z/9873249-33-9.jpg?t=%E4%B8%80%E9%81%93%E4%B8%8D%E7%AD%89%E5%BC%8F%E7%AB%9E%E8%B5%9B%E9%A2%98%E8%AE%BEa%2Cb%2Cc%E2%88%88R%2B%E4%B8%94a%2Bb%2Bc%3D3%2C%E6%B1%82%E8%AF%81%28a%26%23178%3B%2B3b%26%23178%3B%29%2Fab%26%23178%3B%284-ab%29%2B%28b%26%23178%3B%2B3c%26%23178%3B%29%2Fbc%26%23178%3B%284-bc%29%2B%28c%26%23178%3B%2B3a%26%23178%3B%29%2Fca%26%23178%3B%284-ac%29%E2%89%A54)
一道不等式竞赛题设a,b,c∈R+且a+b+c=3,求证(a²+3b²)/ab²(4-ab)+(b²+3c²)/bc²(4-bc)+(c²+3a²)/ca²(4-ac)≥4
一道不等式竞赛题
设a,b,c∈R+且a+b+c=3,求证(a²+3b²)/ab²(4-ab)+(b²+3c²)/bc²(4-bc)+(c²+3a²)/ca²(4-ac)≥4
一道不等式竞赛题设a,b,c∈R+且a+b+c=3,求证(a²+3b²)/ab²(4-ab)+(b²+3c²)/bc²(4-bc)+(c²+3a²)/ca²(4-ac)≥4
首先有(a²+3b²)/(ab²(4-ab)) ≥ (2ab+2b²)/(ab²(4-ab)) = 2(a+b)/(ab(4-ab)).
同理(b²+3c²)/(bc²(4-bc)) ≥ 2(b+c)/(bc(4-bc)),(c²+3a²)/(ca²(4-ca)) ≥ 2(c+a)/(ca(4-ca)).
只需证明:36(a+b)/(ab(4-ab))+36(b+c)/(bc(4-bc))+36(c+a)/(ca(4-ca)) ≥ 72.
而36(a+b)/(ab(4-ab)) = 9(a+b)(1/(ab)+1/(4-ab))
= 9/b+9/a+9a/(4-ab)+9b/(4-ab)
= 9a/(4-ab)+(4-ab)/a+9b/(4-ab)+(4-ab)/b+5/a+5/b+a+b
≥ 6+6+5/a+5/b+a+b
= 12+5(1/a+1/b)+(a+b).
同理36(b+c)/(bc(4-bc)) ≥ 12+5(1/b+1/c)+(b+c),36(c+a)/(ca(4-ca)) ≥ 12+5(1/c+1/a)+(c+a).
于是36(a+b)/(ab(4-ab))+36(b+c)/(bc(4-bc))+36(c+a)/(ca(4-ca))
≥ 36+10(1/a+1/b+1/c)+2(a+b+c)
= 42+10(1/a+1/b+1/c)
而由Cauchy不等式得(a+b+c)(1/a+1/b+1/c) ≥ 9,即1/a+1/b+1/c ≥ 3.
代入即得:36(a+b)/(ab(4-ab))+36(b+c)/(bc(4-bc))+36(c+a)/(ca(4-ca)) ≥ 72.