Question

Cho a, b, c \(\ge\)0; a + b + c = 1. CMR: \(\sqrt[]{a+b}+\sqrt{b+c}+\sqrt{c+1}\le\sqrt{6}\)

Answer 1

Áp dụng BĐT Bunhiakovski

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

\(\le\left(1^2+1^2+1^2\right)\left(a+b+b+c+c+a\right)\)

\(=3.2\left(a+b+c\right)=6\)

Do đó \(VT\le\sqrt{6}\)

Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{a+b}}{1}=\dfrac{\sqrt{b+c}}{1}=\dfrac{\sqrt{c+a}}{1}\\a+b+c=1\end{matrix}\right.\)

\(\Leftrightarrow a=b=c=\dfrac{1}{3}\)