已知正实数 a、b、ca、b、ca、b、c 满足 ab+bc+ca≤1ab+bc+ca\leq1ab+bc+ca≤1. 证明:
a+b+c+3≥8abc(1a2+1)(1b2+1)(1c2+1).a+b+c+\sqrt3\geq8abc(\frac1{a^2+1})(\frac1{b^2+1})(\frac1{c^2+1}).a+b+c+3≥8abc(a2+11)(b2+11)(c2+11).
\qquad证明\qquad由均值不等式得
\qquad a2+1≥a2+ab+bc+ca≥4a2⋅ab⋅bc⋅ca4=4abc.a^2+1\geq a^2+ab+bc+ca\geq4\sqrt[4]{a^2\cdot ab\cdot bc\cdot ca}=4a\sqrt{bc}.a2+1≥a2+ab+bc+ca≥44a2⋅ab⋅bc⋅ca=4abc.
\qquad于是, 2bc≥8abca2+12\sqrt{bc}\geq\frac{8abc}{a^2+1}2bc≥a2+18abc
\qquad同理,2ca≥8abcb2+1,2ab≥8abcc2+12\sqrt{ca}\geq\frac{8abc}{b^2+1},2\sqrt{ab}\geq\frac{8abc}{c^2+1}2ca≥b2+18abc,2ab≥c2+18abc
\qquad只需证:a+b+c+3≥2(bc+ca+ab).a+b+c+\sqrt3\geq2(\sqrt{bc}+\sqrt{ca}+\sqrt{ab}).a+b+c+3≥2(bc+ca+ab).
\qquad由柯西不等式知
\qquad3≥1+1+1⋅ab+bc+ca≥ab+bc+ca\sqrt3\geq\sqrt{1+1+1}\cdot\sqrt{ab+bc+ca}\geq\sqrt{ab}+\sqrt{bc}+\sqrt{ca}3≥1+1+1⋅ab+bc+ca≥ab+bc+ca
\qquad而 a+b+c≥ab+bc+caa+b+c\geq\sqrt{ab}+\sqrt{bc}+\sqrt{ca}a+b+c≥ab+bc+ca
⇔(a−b)2+(b−c)2+(c−a)2≥0.\Leftrightarrow(\sqrt a-\sqrt b)^2+(\sqrt b-\sqrt c)^2+(\sqrt c-\sqrt a)^2\geq0.⇔(a−b)2+(b−c)2+(c−a)2≥0.
\qquad从而,所证不等式成立.