1, $\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(a+c)^{2}}\geq \frac{3\sqrt[3]{3abc(a+b+c)}(a+b+c)^{2}}{4(ab+bc+ca)^{2}}$
2,Cho: a+b+c=$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Chứng minh rằng:
$(ab+bc+ca)(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^{2}\geq 27$