Question

cho a b c là các số thực dương thỏa mãn a+b+c=1

cmr \(\sqrt{5a+1}+\sqrt{5b+1}+\sqrt{5c+1}< =2\sqrt{6}\)

Answer 1

A=\(\sqrt{5a+1}+\sqrt{5b+1}+\sqrt{5c+1}\)(\(A\ge0\))

<=> \(A^2=\left(\sqrt{5a+1}+\sqrt{5b+1}+\sqrt{5c+1}\right)^2\)

Áp dụng bđt bunhiacopski có:

\(\left(1.\sqrt{5a+1}+1.\sqrt{5b+1}+1.\sqrt{5c+1}\right)^2\le\left(1+1+1\right)\left(5a+1+5b+1+5c+1\right)\)

<=> \(A^2\le3\left(5a+5b+5c+3\right)=3.\left[5\left(a+b+c\right)+3\right]=3\left(5.1+3\right)=24\)(do a+b+c=1)

<=> \(A\le2\sqrt{6}\)

Dấu"=" xảy ra <=> \(a=b=c=\frac{1}{3}\)

Vậy \(A\le2\sqrt{6}\)