\(\sqrt{a\left(3b+c\right)}+\sqrt{b\left(3c+a\right)}+\sqrt{c\left(3a+b\right)}=\dfrac{\sqrt{4a\left(3b+c\right)}=\sqrt{4b\left(3c+a\right)}+\sqrt{4c\left(3a+b\right)}}{2}\le\dfrac{\left(4a+3b+c\right)+\left(4b+3c+a\right)+\left(4c+3a+b\right)}{4}\)\(=\dfrac{8\left(a+b+c\right)}{4}=2\left(a+b+c\right)\)
Dấu "=" xảy ra <=> a = b = c
Theo BĐT Cô - Si ta có :
\(\left\{{}\begin{matrix}\sqrt{a\left(3b+c\right)}\le\dfrac{a+3b+c}{2}\\\sqrt{b\left(3c+a\right)}\le\dfrac{b+3c+a}{2}\\\sqrt{c\left(3a+b\right)}\le\dfrac{c+3a+b}{2}\end{matrix}\right.\)
Cộng từng vế của BĐT ta được :
\(\sqrt{a\left(3b+c\right)}+\sqrt{b\left(3c+a\right)}+\sqrt{c\left(3a+b\right)}\le\dfrac{5\left(a+b+c\right)}{2}=2,5\left(a+b+c\right)\)
Chịu @@
áp dụng bất đẳng thức \(Bunhiacopxki\) ta có :
\(\sqrt{a\left(3b+c\right)}+\sqrt{b\left(3c+a\right)}+\sqrt{c\left(3a+b\right)}\le\sqrt{\left(a+b+c\right)\left(4a+4b+4c\right)}\)
\(=2\left(a+b+c\right)\left(đpcm\right)\)
dấu "=" xảy ra khi \(a=b=c\)
@DƯƠNG PHAN KHÁNH DƯƠNG @Mysterious Person
Lời giải
Áp dụng bđt Bunyakovsky:
\(VT^2=\left(\sqrt{a\left(3b+c\right)}+\sqrt{b\left(3c+a\right)}+\sqrt{c\left(3a+b\right)}\right)^2\)
\(\le\left(1^2+1^2+1^2\right)\left(3ab+ac+3bc+ab+3ac+bc\right)\)
\(=12\left(ab+bc+ac\right)\le\dfrac{12\left(a+b+c\right)^2}{3}=4\left(a+b+c\right)^2\)(Bđt AM-GM)
\(\Leftrightarrow VT\le2\left(a+b+c\right)\). \("="\Leftrightarrow a=b=c\)
