Question

Cho các số a,b,c dương thỏa mãn abc=1. Chứng minh rằng \(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+a+b+c\ge4\left(\dfrac{a}{a^4+1}+\dfrac{b}{b^4+1}+\dfrac{c}{c^4+1}\right)\)

Answer 1

\(\dfrac{1}{a^3}+a\ge2\sqrt{\dfrac{a}{a^3}}=\dfrac{2}{a}\) ; \(\dfrac{1}{b^3}+b\ge\dfrac{2}{b}\) ; \(\dfrac{1}{c^3}+c\ge\dfrac{2}{c}\)

\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+a+b+c\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) (1)

Lại có \(\dfrac{4a}{a^4+1}\le\dfrac{4a}{2\sqrt{a^4}}=\dfrac{4a}{2a^2}=\dfrac{2}{a}\)

Tương tự \(\dfrac{4b}{b^4+1}\le\dfrac{2}{b}\) ; \(\dfrac{4c}{c^4+1}\le\dfrac{2}{c}\)

\(\Rightarrow4\left(\dfrac{a}{a^4+1}+\dfrac{b}{b^4+1}+\dfrac{c}{c^4+1}\right)\le2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) (2)

Từ (1),(2)\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+a+b+c\ge4\left(\dfrac{a}{a^4+1}+\dfrac{b}{b^4+1}+\dfrac{c}{c^4+1}\right)\)

Dấu "=" xảy ra khi a=b=c=1