Question

cho x,y>0 thỏa mãn \(2x^2+3y^2=4\)

Cmr: \(x+2y\le\sqrt{\frac{22}{3}}\)

Answer 1

Áp dụng BĐT Bunhicopxki:

\(\left(\sqrt{\frac{1}{2}}^2+\sqrt{\frac{4}{3}}^2\right)\left(\left(\sqrt{2}x\right)^2+\left(\sqrt{3}y\right)^2\right)\ge\left(x+2y\right)^2\)

\(\Leftrightarrow\frac{11}{6}\left(2x^2+3y^2\right)\ge\left(x+2y\right)^2\)

\(\Leftrightarrow\frac{44}{6}=\frac{22}{3}\ge\left(x+2y\right)^2\)(1)

Do x, y > 0 nên x + 2y > 0 do đó từ (1) suy ra \(x+2y\le\sqrt{\frac{22}{3}}\)(đpcm)