Bài toán (Nguyễn Đình Thi) Chứng minh rằng với mọi $a,b,c\in\mathbb{R}$ thì $$\left(\dfrac{a^2+bc}{a^2-bc}\right)^2+\left(\dfrac{b^2+ca}{b^2-ca}\right)^2+\left(\dfrac{c^2+ab}{c^2-ab}\right)^2\ge 2$$  
Lời giải: 
Đặt $x=\dfrac{a^2+bc}{a^2-bc}, y=\dfrac{b^2+ca}{b^2-ca},z=\dfrac{c^2+ab}{c^2-ab}$ Khi đó, ta có: $$(x+1)(y+1)(z+1)=(x-1)(y-1)(z-1)\Leftrightarrow xy+yz+zx=-1$$ Áp dụng BDT $$x^2+y^2+z^2\ge -2(xy+yz+zx)$$ Ta có $$\left(\dfrac{a^2+bc}{a^2-bc}\right)^2+\left(\dfrac{b^2+ca}{b^2-ca}\right)^2+\left(\dfrac{c^2+ab}{c^2-ab}\right)^2\ge 2$$ Ta có đpcm.


Problem (Nguyen Dinh Thi) Let $a, b, c$ be real numbers. Prove that:  $$\left(\dfrac{a^2+bc}{a^2-bc}\right)^2+\left(\dfrac{b^2+ca}{b^2-ca}\right)^2+\left(\dfrac{c^2+ab}{c^2-ab}\right)^2\ge 2$$ Proof: 
Let $x=\dfrac{a^2+bc}{a^2-bc}, y=\dfrac{b^2+ca}{b^2-ca},z=\dfrac{c^2+ab}{c^2-ab}$ Then,  we have:  $$(x+1)(y+1)(z+1)=(x-1)(y-1)(z-1)\Leftrightarrow xy+yz+zx=-1$$ Using this inequality:  $$x^2+y^2+z^2\ge -2(xy+yz+zx)$$ We have:  $$\left(\dfrac{a^2+bc}{a^2-bc}\right)^2+\left(\dfrac{b^2+ca}{b^2-ca}\right)^2+\left(\dfrac{c^2+ab}{c^2-ab}\right)^2\ge 2\;\;\text{(Q.E.D)}$$