Cho $a, b, c > 0$ thỏa mãn $a+b+c=1$ . CMR: $\frac{bc}{\sqrt{a+bc}}+\frac{ca}{\sqrt{b+ca}}+ \frac{ab}{\sqrt{c+ab}} \le \frac{1}{2}$

Biến đổi một chút, ta có: $$\dfrac{bc}{\sqrt{a+bc}}=\dfrac{bc}{\sqrt{a(a+b+c)+bc}}=\sqrt{\dfrac{bc}{a+b}}.\sqrt{\dfrac{bc}{c+a}} \leq \dfrac{1}{2} \left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right)$$

Tương tự: $$\dfrac{ca}{\sqrt{b+ca}}\leq \dfrac{1}{2} \left(\dfrac{ca}{a+b}+\dfrac{ac}{b+c}\right);\,\,\,\,\dfrac{ab}{\sqrt{c+ab}}\leq \dfrac{1}{2} \left(\dfrac{ab}{a+c}+\dfrac{ab}{a+b}\right)$$

Cộng ba bất đẳng thức trên lại theo vế, ta có:
$\dfrac{bc}{\sqrt{a+bc}}+\dfrac{ca}{\sqrt{b+ca}}+ \dfrac{ab}{\sqrt{c+ab}} \leq \dfrac{1}{2} \left(a+b+c\right)=\dfrac{1}{2}$