Ta có \(B=\sqrt{x+3}+\sqrt{5-x}\Leftrightarrow B^2=x+3+5-x+2\sqrt{\left(x+3\right)\left(5-x\right)}=8+2\sqrt{\left(x+3\right)\left(5-x\right)}\) Ta có \(\sqrt{\left(x+3\right)\left(5-x\right)}\ge0\Leftrightarrow2\sqrt{\left(x+3\right)\left(5-x\right)}\ge0\Leftrightarrow8+2\sqrt{\left(x+3\right)\left(5-x\right)}\ge8\Leftrightarrow B^2\ge8\Leftrightarrow B\ge2\sqrt{2}\)Vậy \(2\sqrt{2}\le B\)(1)
Áp dụng bđt Bunhia copski ta có
\(B^2=\left(\sqrt{x+3}+\sqrt{5-x}\right)^2=\left(\sqrt{x+3}.1+\sqrt{5-x}.1\right)^2\le\left[\left(\sqrt{x+3}\right)^2+\left(\sqrt{5-x}\right)^2\right]\left(1^2+1^2\right)=8.2=16\Leftrightarrow B^2\le16\Leftrightarrow B\le4\)(2)
Từ (1),(2)\(\Rightarrow2\sqrt{2}\le B\le4\)
