\(GT\Leftrightarrow\frac{1}{1+a}-1+\frac{1}{1+b}-1+\frac{1}{1+c}-1+\frac{1}{1+d}-1\)\(\ge3-4\)
\(\Rightarrow\frac{-a}{1+a}+\frac{-b}{1+b}+\frac{-c}{1+c}+\frac{-d}{1+d}\ge-1\)
\(\Rightarrow\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}\le1\)
\(\Rightarrow\frac{a\left(1+b\right)+b\left(1+a\right)}{\left(1+a\right)\left(1+b\right)}+\frac{c\left(1+d\right)+d\left(1+c\right)}{\left(1+c\right)\left(1+d\right)}\le1\)
\(\Rightarrow\frac{a+2ab+b}{1+a+b+ab}+\frac{c+2cd+d}{1+c+d+cd}\le1\)
Áp dụng BĐT Cô - si , ta có:
\(1\ge\frac{2\sqrt{ab}+2ab}{1+2\sqrt{ab}+ab}+\frac{2\sqrt{cd}+2cd}{1+2\sqrt{cd}+cd}=\frac{2\sqrt{ab}}{1+\sqrt{ab}}+\frac{2\sqrt{cd}}{1+\sqrt{cd}}\)
\(\Rightarrow1\ge2\left[2\sqrt{\frac{\sqrt{abcd}}{1+\sqrt{ab}+\sqrt{cd}+\sqrt{abcd}}}\right]\)\(=4.\frac{\sqrt[4]{abcd}}{1+\sqrt{ab}+\sqrt{cd}+\sqrt{abcd}}\)
\(\Rightarrow1\ge\frac{4\sqrt[4]{abcd}}{1+2\sqrt[4]{abcd}+\sqrt{abcd}}=\frac{4\sqrt[4]{abcd}}{\sqrt{\left(1+\sqrt[4]{abcd}\right)^2}}\)
\(\Rightarrow4\sqrt[4]{abcd}\le\sqrt{\left(1+\sqrt[4]{abcd}\right)^2}\)
\(\Rightarrow4\sqrt[4]{abcd}\le1+\sqrt[4]{abcd}\)(vì a,b,c,d dương)
\(\Rightarrow3\sqrt[4]{abcd}\le1\)
\(\Rightarrow\sqrt[4]{abcd}\le\frac{1}{3}\)
\(\Rightarrow abcd\le\frac{1}{81}\)
(Dấu "="\(\Leftrightarrow a=b=c=d=\frac{1}{3}\))
Coll boy ! Bài này dòng 5 em áp dụng bất đẳng thức cô-si như vậy là chưa đúng nhé! Em kiểm tra lại mẫu trái dấu em nhé!
Nguyễn Linh Chia,b,c,d dương mà cô. Cô có thể nói cho em hiểu không ạ, chưa thông lắm??? )):
Kiệt sai r.Ý Cô Chi là bạn cosi dưới mẫu thì nó sẽ bị ngược dấu.
Ta có:
My solution
\(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}\ge3\)
\(\Leftrightarrow\frac{1}{1+a}\ge\left(1-\frac{1}{1+b}\right)+\left(1-\frac{1}{1+c}\right)+\left(1-\frac{1}{1+d}\right)\)
\(\Leftrightarrow\frac{1}{1+a}\ge\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}\ge3\sqrt[3]{\frac{bcd}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}>0\) ( cô si )
Tương tự:
\(\frac{1}{1+b}\ge3\sqrt[3]{\frac{cda}{\left(1+a\right)\left(1+c\right)\left(1+d\right)}}>0\)
\(\frac{1}{1+c}\ge3\sqrt[3]{\frac{abd}{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}>0\)
\(\frac{1}{1+d}\ge3\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}>0\)
Nhân các vế BĐT lại ta được:
\(\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\ge81\sqrt[3]{\frac{a^3b^3c^3d^3}{\left(1+a\right)^3\left(1+b\right)^3\left(1+c\right)^3\left(1+d\right)^3}}\)
\(\Leftrightarrow\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\ge81\cdot\frac{abcd}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\)
\(\Leftrightarrow abcd\le\frac{1}{81}\)