Question

CMR : Với mọi x và y ta luôn có

\(\sqrt{\frac{x^2+4y^2}{2}}+\sqrt{\frac{x^2+2xy+4y^2}{3}}\ge x+2y\)

Answer 1

Áp dụng BĐT \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\):

\(VT=\sqrt{\frac{x^2+\left(2y\right)^2}{2}}+\sqrt{\frac{\left(\frac{x}{2}-y\right)^2+3\left(\frac{x}{2}+y\right)^2}{3}}\)

\(VT\ge\sqrt{\frac{\left(x+2y\right)^2}{4}}+\sqrt{\frac{3\left(\frac{x}{2}+y\right)^2}{3}}\)

\(VT\ge\left|\frac{x+2y}{2}\right|+\left|\frac{x+2y}{2}\right|=\left|x+2y\right|\ge x+2y\) (đpcm)

Dấu "=" xảy ra khi \(x=2y\ge0\)