Question

chứng minh rằng √(x^2+2x+5)  +√(2x^2+4x+6) ≥ 4

Answer 1

\(VT=\sqrt{x^2+2x+5}+\sqrt{2x^2+4x+6}\)

\(=\sqrt{x^2+2x+1+4}+\sqrt{2x^2+4x+2+4}\)

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

Dễ thấy: \(\hept{\begin{cases}\sqrt{\left(x+1\right)^2}\ge0\\\sqrt{2\left(x+1\right)^2}\ge0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\sqrt{\left(x+1\right)^2+4}\ge\sqrt{4}=2\\\sqrt{2\left(x+1\right)^2+4}\ge\sqrt{4}=2\end{cases}}\)

\(\Rightarrow\sqrt{\left(x+1\right)^2+4}+\sqrt{2\left(x+1\right)^2+4}\ge2+2=4\)

Đẳng thức xảy ra khi \(x=-1\)