Question

[Phạm hải vương]

Cho a , b , c dương :

Chứng minh rằng : \(\left(1+\frac{1}{a}\right)^4+\left(1+\frac{1}{b}\right)^4+\left(1+\frac{1}{c}\right)^4\ge3\left(1+\frac{3}{2+abc}\right)^4\)

Answer 1

[Agatsuma Zenitsu]

Áp dụng BĐT Cô-si cho 3 số dương ta có:

\(\left(1+\frac{1}{a}\right)^4+\left(1+\frac{1}{b}\right)^4+\left(1+\frac{1}{c}\right)^4\ge3\left(\sqrt[3]{\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)}\right)^4\)

Ta chứng minh: \(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\ge\left(1+\frac{3}{2+abc}\right)^3\left(1\right)\)

Theo BĐT Cô - si ta có:

\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}\)

\(\ge1+\frac{3}{\sqrt[3]{abc}}+\frac{3}{\sqrt[3]{\left(abc\right)^2}}+\frac{1}{abc}=\left(1+\frac{1}{\sqrt[3]{abc}}\right)^3\ge\left(1+\frac{3}{2+abc}\right)^3\)

(Vì \(abc+2=abc+1+1\ge3\sqrt[3]{abc}\))

Vậy \(\left(1\right)\) được chứng minh \(\Rightarrow BĐT\) đúng \(\forall a,b,c>0\)

Đẳng thức xảy ra \(\Leftrightarrow a=b=c=1\)

Answer 2

[Kudo Shinichi]

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow VT\ge3\sqrt[3]{\left[\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\right]^4}\)

\(\Rightarrow VT\ge3\left(\sqrt[3]{1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}}\right)^4\left(1\right)\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\hept{\begin{cases}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\\\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge3\sqrt[3]{\frac{1}{a^2b^2c^2}}\end{cases}}\)

\(\Rightarrow1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}\ge1+3\sqrt[3]{\frac{1}{abc}}\)

\(+3\sqrt[3]{\frac{1}{a^2b^2c^2}}+\frac{1}{abc}\)

\(\Rightarrow1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}\ge\left(1+\frac{1}{\sqrt[3]{abc}}\right)^3\)

\(\Rightarrow3\left(\sqrt[3]{1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}}\right)^4\)

\(\ge3\left(1+\frac{1}{\sqrt[3]{abc}}\right)^4\)

\(\left(2\right)\)

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow\sqrt[3]{abc}\le\frac{abc+1+1}{3}=\frac{abc+2}{3}\)

\(\Rightarrow1+\frac{1}{\sqrt[3]{abc}}\ge1+\frac{3}{abc+2}\)

\(\Rightarrow3\left(1+\frac{1}{\sqrt[3]{abc}}\right)^4\ge3\left(1+\frac{3}{abc+2}\right)^4\left(3\right)\)

Từ (1) , (2) và (3) 

\(\Rightarrow VT\ge3\left(1+\frac{3}{abc+2}\right)^4\)

\(\Leftrightarrow\left(1+\frac{1}{a}\right)^4+\left(1+\frac{1}{b}\right)^4+\left(1+\frac{1}{c}\right)^4\ge3\left(1+\frac{3}{2+abc}\right)^4\left(đpcm\right)\)

Chúc bạn học tốt !!!