Question

Cho a, b, c \(\ge\dfrac{-3}{4}\) và a + b + c + d = 3. CMR: \(\sqrt{4a+3}+\sqrt{4b+3}+\sqrt{4c+3}\le3\sqrt{7}\)

Answer 1

d đâu ra vậy bạn ?

Answer 2

Đặt \(A=\sqrt{4a+3}+\sqrt{4b+3}+\sqrt{4c+3}\Rightarrow A^2=\left(\sqrt{4a+3}+\sqrt{4b+3}+\sqrt{4c+3}\right)^2\)

Áp dụng BĐT Bu - nhi - a - cốp - xki ta có :

\(A^2=\left(\sqrt{4a+3}+\sqrt{4b+3}+\sqrt{4c+3}\right)^2\le\left(1^2+1^2+1^2\right)\left(4a+3+4b+3+4c+3\right)=3\left[4\left(a+b+c\right)+9\right]=3\left(12+9\right)=63\)

\(\Rightarrow A=\sqrt{4a+3}+\sqrt{4b+3}+\sqrt{4c+3}\le\sqrt{63}=3\sqrt{7}\)

Dấu \("="\) xảy ra khi \(a=b=c=1\)