Question

Cho abcd = 1. Chứng minh rằng \(\frac{1}{1+ab+bc+ca}+\frac{1}{1+bc+cd+db}+\frac{1}{1+cd+da+ac}+\frac{1}{1+da+ab+bd}\le1\)

Answer 1

Xét \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ac}}=\sqrt{d}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)

\(\Rightarrow ab+bc+ac\ge\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{d}}\) và \(\frac{1}{1+ab+bc+ac}\le\frac{\sqrt{d}}{\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}}\)

Tương tự : \(\frac{1}{1+bc+cd+da}\le\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}}\)

\(\frac{1}{1+cd+da+ac}\le\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}}\)

\(\frac{1}{1+da+ab+bd}\le\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}}\)

Cộng theo vế ta được đpcm.