Question

\(Cho\sqrt{1+x}+\sqrt{1+y}=2\sqrt{1+a}\)

\(CMR:x+y\ge2a\)

Answer 1

\(\left(2\sqrt{1+a}\right)^2=4\left(1+a\right)=\left(\sqrt{1+x}+\sqrt{1+y}\right)^2\le2\left(x+y+2\right)\)

\(\Leftrightarrow\)\(x+y\ge2a\)

Answer 2

Áp dụng bđt Bunyakovsky: \(\left(\sqrt{1+x}+\sqrt{1+y}\right)^2\le2\left(x+y+2\right)\)

\(\Rightarrow4\left(a+1\right)\le2\left(x+y+2\right)\Leftrightarrow4a\le2\left(x+y\right)\Leftrightarrow x+y\ge2a\)