Question

Cho \(a,b,c\) thỏa mãn \(\left|a\right|,\left|b\right|,\left|c\right|< 1\) và \(ab+bc+ca=2\). Chứng minh : 

\(P=\dfrac{a^2}{1-b^2}+\dfrac{b^2}{1-c^2}+\dfrac{c^2}{1-a^2}\ge6\).

Answer 1

\(a^2+b^2+c^2\ge ab+bc+ca=2\)

Áp dụng BĐT C-S:

\(P\ge\dfrac{\left(a+b+c\right)^2}{3-\left(a^2+b^2+c^2\right)}=\dfrac{a^2+b^2+c^2+4}{3-\left(a^2+b^2+c^2\right)}\)

Đặt \(a^2+b^2+c^2=x\)

Ta cần c/m: \(\dfrac{x+4}{3-x}\ge6\Leftrightarrow x+4\ge18-6x\)

\(\Leftrightarrow x\ge2\) (đúng)

Dấu = xảy ra khi \(a=b=c=\pm\sqrt{\dfrac{2}{3}}\)