Question

[Anh Triệu Quốc]

a,b,c>0; a+b+c=3

CMR\(a\sqrt{b+3c}\)+ \(b\sqrt{c+3a}\)+\(c\sqrt{a+3b}\)<=6

Answer 1

[Đoàn Phương Liên]

Áp dụng BĐT AM-GM cho 2 số dương, ta có:

\(\left(b+3c\right)+4\ge2\sqrt{4\left(b+3c\right)}=4\sqrt{b+3c}\\ \)

\(\Rightarrow\sqrt{b+3c}\le\frac{b+3c+4}{4}\)

\(\Rightarrow a\sqrt{b+3c}\le\frac{ab+3ac+4a}{4}\)

Tương tự ta có \(b\sqrt{c+3a}\le\frac{bc+3ab+4b}{4}\)

\(c\sqrt{a+3b}\le\frac{ac+3bc+4c}{4}\)

\(\Rightarrow a\sqrt{b+3c}+b\sqrt{c+3a}+c\sqrt{a+3b}\le\)\(\frac{4\left(ab+bc+ca\right)+4\left(a+b+c\right)}{4}\)\(=\frac{4\left(ab+bc+ac\right)+12}{4}\)

Ta có bổ đề:3(ab+bc=ca) \(\le\)(a+b+c)^2 => 3(ab+bc+ca) \(\le9\)=> \(\text{(ab+bc+ca)}\le3\)

=>\(a\sqrt{b+3c}+b\sqrt{c+3a}+c\sqrt{a+3b}\le\)\(\frac{4.3+12}{4}=6\left(đpcm\right)\)

Dấu "=" xảy ra <=>a=b=c=1