Question

Cho các số thực dương a, b, c thỏa mãn \(a^2+b^2+c^2=1\). Chúng minh:

\(\frac{a}{\sqrt{1-a^2}}+\frac{b}{\sqrt{1-b^2}}+\frac{c}{\sqrt{1-c^2}}>2\)

Answer 1

\(\frac{a}{\sqrt{1-a^2}}=\frac{a^2}{a\sqrt{1-a^2}}\ge\frac{a^2}{\frac{a^2+1-a^2}{2}}=2a^2\)

Tương tự \(\frac{b}{\sqrt{1-b^2}}\ge2b^2\); \(\frac{c}{\sqrt{1-c^2}}\ge2c^2\)

Cộng vế với vế: \(\frac{a}{\sqrt{1-a^2}}+\frac{b}{\sqrt{1-b^2}}+\frac{c}{\sqrt{1-c^2}}\ge2\left(a^2+b^2+c^2\right)=2\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a^2=1-a^2\\b^2=1-b^2\\c^2=1-c^2\end{matrix}\right.\) \(\Rightarrow a^2=b^2=c^2=\frac{1}{2}\)

\(\Rightarrow a^2+b^2+c^2=\frac{3}{2}\) trái giả thiết \(a^2+b^2+c^2=1\) nên dấu "=" ko xảy ra

Vậy \(\frac{a}{\sqrt{1-a^2}}+\frac{b}{\sqrt{1-b^2}}+\frac{c}{\sqrt{1-c^2}}>2\)