Question

Cho a, b, c là các số thực không âm. CM nếu \(\sqrt{1+b}+\sqrt{1+c}=2\sqrt{1+a}\) thì \(b+c\ge2a\)

Answer 1

\(\sqrt{1+b}+\sqrt{1+c}=2\sqrt{1+a}\) (1)

⇒ \(b+c+2+2\sqrt{\left(1+b\right)\left(1+c\right)}=4a+4\)

⇒ \(b+c=4a+2-2\sqrt{\left(1+b\right)\left(1+c\right)}\)

Từ (1) ta lại có: \(2\sqrt{1+a}=\sqrt{1+b}+\sqrt{1+c}\ge2\sqrt[4]{\left(1+b\right)\left(1+c\right)}\)

⇒ \(1+a\ge\sqrt{\left(1+b\right)\left(1+c\right)}\) ⇒

\(b+c=4a+2-2\sqrt{\left(1+b\right)\left(1+c\right)}\ge4a+2-2\left(1+a\right)=2a\)

Vậy \(b+c\ge2a\), "=" xảy ra khi \(b=c=a\)