Áp dụng BĐT C-S:
$$\frac{ab}{\sqrt{\frac{ab+bc}{2}}}+\frac{bc}{\sqrt{\frac{bc+ca}{2}}}+\frac{ca}{\sqrt{\frac{ca+ab}{2}}}\leq \sqrt{2[ab(a+b)+bc(b+c)+ca(c+a)]\left[\frac{a}{(a+b)(a+c)}+\frac{b}{(b+c)(b+a)}+\frac{c}{(c+a)(c+b)}\right]}=2\sqrt{(bc+ca+ab)\left[\frac{bc}{(a+b)(a+c)}+\frac{ca}{(b+c)(b+a)}+\frac{ab}{(c+a)(c+b)}\right]}.$$
Với mọi số thực dương $a,b,c,x,y,z$, ta có:
$$2\sqrt{(bc+ca+ab)(yz+zx+xy)}+ax+by+cz=\sqrt{2(bc+ca+ab)\cdot 2(yz+zx+xy)}+ax+by+cz\leq \sqrt{[2(bc+ca+ab)+a^{2}+b^{2}+c^{2}][2(yz+zx+xy)+x^{2}+y^{2}+z^{2}]}=(a+b+c)(x+y+z)$$
$\Rightarrow 2\sqrt{(bc+ca+ab)(yz+zx+xy)}\leq x(b+c)+y(c+a)+z(a+b).$
Chọn $x=\frac{a}{b+c},y=\frac{b}{c+a},z=\frac{c}{a+b}$ ta có
$$2\sqrt{(bc+ca+ab)\left[\frac{bc}{(a+b)(a+c)}+\frac{ca}{(b+c)(b+a)}+\frac{ab}{(c+a)(c+b)}\right]}\leq a+b+c=3$$
$$\Rightarrow \frac{ab}{\sqrt{\frac{ab+bc}{2}}}+\frac{bc}{\sqrt{\frac{bc+ca}{2}}}+\frac{ca}{\sqrt{\frac{ca+ab}{2}}}\leq 3,$$
hay $$\frac{ab}{\sqrt{ab+bc}}+\frac{bc}{\sqrt{bc+ca}}+\frac{ca}{\sqrt{ca+ab}}\leq \frac{3}{\sqrt{2}}.$$
Đẳng thức xảy ra chỉ khi $a=b=c=1$. $\square$
lời giải hay quá. cảm ơn bạn