\(Q=\sqrt{x+3}+\sqrt{10-x}\)
\(\Leftrightarrow Q^2=\left(\sqrt{x+3}+\sqrt{10-x}\right)^2\le\left(1^2+1^2\right)\left[\left(\sqrt{x+3}\right)^2+\left(\sqrt{10-x}\right)^2\right]\)
\(\Leftrightarrow Q^2\le2\left(x+3+10-x\right)=2.13=26\)
\(\Leftrightarrow Q\le\sqrt{26}\)
\(maxQ=\sqrt{26}\Leftrightarrow x+3=10-x\Leftrightarrow x=\dfrac{7}{2}\)
Áp dụng BĐT Bunhiacopski:
\(Q=\sqrt{x+3}+\sqrt{10-x}\\ \Leftrightarrow Q^2=\left(\sqrt{x+3}+\sqrt{10-x}\right)^2\le\left(1^2+1^2\right)\left(x+3+10-x\right)=2\cdot13=26\\ \Leftrightarrow Q\le\sqrt{26}\\ Q_{max}=\sqrt{26}\Leftrightarrow x+3=10-x\Leftrightarrow x=\dfrac{7}{2}\)
