Question

Cho các số thực: 0\(\le\)a\(\le\)1; 0\(\le\)b\(\le\)1; 0\(\le\)c\(\le\)1 thoả mãn:

\(a\sqrt{1-b^2}+b\sqrt{1-c^2}+c\sqrt{1-a^2}=\dfrac{3}{2}\)

Chứng minh: \(a^2+b^2+c^2=\dfrac{3}{2}\)

Answer 1

Áp dụng BĐT cosi:

\(a\sqrt{1-b^2}=\sqrt{a^2\left(1-b^2\right)}\le\dfrac{a^2+1-b^2}{2}\)

Tương tự cx có: \(b\sqrt{1-c^2}\le\dfrac{b^2+1-c^2}{2}\)

\(c\sqrt{1-a^2}\le\dfrac{c^2+1-a^2}{2}\)

Cộng vế với vế \(\Rightarrow VT\le\dfrac{3}{2}\)

Dấu = xảy ra <=> \(\left\{{}\begin{matrix}a^2=1-b^2\\b^2=1-c^2\\c^2=1-a^2\end{matrix}\right.\) \(\Leftrightarrow a^2+b^2+c^2=3-\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow a^2+b^2+c^2=\dfrac{3}{2}\) (đpcm)