Question

mọi người giúp mình bài này nha

thanks nhiều

0<x<1; 0<y<1

C/m: \(x+y+x\sqrt{1-y^2}+y\sqrt{1-x^2}< =\dfrac{3\sqrt{3}}{2}\)

Answer 1

Áp dụng BĐT Bunyakovsky:

\(VT^2=\left(\sqrt{2}x.\dfrac{1}{\sqrt{2}}+\sqrt{2}y.\dfrac{1}{\sqrt{2}}+x.\sqrt{1-y^2}+y.\sqrt{1-x^2}\right)^2\)

\(\le\left(2x^2+2y^2+x^2+y^2\right)\left(\dfrac{1}{2}+\dfrac{1}{2}+1-x^2+1-y^2\right)\)

\(=3\left(x^2+y^2\right)\left[3-\left(x^2+y^2\right)\right]\le\dfrac{3}{4}.\left(x^2+y^2+3-x^2-y^2\right)^2=\dfrac{3}{4}.9=\dfrac{27}{4}\)

\(\Rightarrow VT\le\dfrac{3\sqrt{3}}{2}\)

Dấu = xảy ra khi \(x=y=\dfrac{\sqrt{3}}{2}\)