Question

[Lil Shroud]

Cho a, b, c > 0 có a + b + c = 3. Chứng minh: \(\sqrt{a\left(b+c+2\right)}+\sqrt{b\left(c+a+2\right)}+\sqrt{c\left(a+b+2\right)}\le6\)

Answer 1

[Nguyễn Hoàng Minh]

\(a+b+c=3\\ \Leftrightarrow a\left(b+c+2\right)=ab+ac+a+b+c+1=\left(a+1\right)\left(b+c+1\right)\)

Tương tự:

\(b\left(c+a+2\right)=\left(b+1\right)\left(a+c+1\right)\\ c\left(a+b+2\right)=\left(c+1\right)\left(a+b+1\right)\)

Áp dụng BĐT cosi:

\(\left\{{}\begin{matrix}\left(a+1\right)\left(b+c+1\right)\le\dfrac{\left(a+1+b+c+1\right)^2}{2}=\dfrac{2^2}{2}=2\\\left(b+1\right)\left(a+c+1\right)\le\dfrac{\left(b+1+a+c+1\right)^2}{2}=\dfrac{2^2}{2}=2\\\left(c+1\right)\left(a+b+1\right)\le\dfrac{\left(c+1+a+b+1\right)^2}{2}=\dfrac{2^2}{2}=2\end{matrix}\right.\)

Cộng vế theo vế 2 BĐT trên:

\(\Leftrightarrow\sqrt{a\left(b+c+2\right)}+\sqrt{b\left(c+a+2\right)}+\sqrt{c\left(a+b+2\right)}\le2+2+2=6\)

Dấu \("="\Leftrightarrow a=b=c=1\)

Answer 2

[Lil Shroud]

anh oi, tại sao chỗ a(b + c + 2) = ab + ac + a + b + c + 1 được ạ? :<