Question

Cho a, b, c ≥ 0 không có hai số nào đồng thời bằng 0. Chứng minh rằng \(\dfrac{bc}{\left(a+\sqrt{bc}\right)^2}+\dfrac{ca}{\left(b+\sqrt{ca}\right)^2}+\dfrac{ab}{\left(c+\sqrt{ab}\right)^2}+\dfrac{2abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge1\)

Answer 1

Bunhiacopxki: \(\left(a+b\right)\left(a+c\right)\ge\left(a+\sqrt{bc}\right)^2\Rightarrow\dfrac{bc}{\left(a+\sqrt{bc}\right)^2}\ge\dfrac{bc}{\left(a+b\right)\left(a+c\right)}\)

Tương tự và cộng lại:

\(\Rightarrow VT\ge\dfrac{bc}{\left(a+b\right)\left(a+c\right)}+\dfrac{ca}{\left(b+a\right)\left(b+c\right)}+\dfrac{ab}{\left(c+a\right)\left(c+b\right)}+\dfrac{2abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(\Rightarrow VT\ge\dfrac{ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=1\) (đpcm)