1. ## prove

i have a problem to prove this..

Let a,b,c be positive real numbers such that abc=1, prove that

1/(a^3*(b+c))+1/(b^3*(c+a))+1/(c^3*(a+b))>=3/2.

can anyone solve this problem? thank you

2. Let $a=\frac{1}{x}, \ b=\frac{1}{y}, \ c=\frac{1}{z}\Rightarrow xyz=1$

$\frac{1}{a^3(b+c)}=\frac{1}{\frac{1}{x^3}\left(\fr ac{1}{y}+\frac{1}{z}\right)}=\frac{x^3yz}{y+z}=\fr ac{x^2}{y+z}$

The inequality becomes

$\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge q\frac{3}{2}$

By Cauchy-Schwarz

$[(y+z)+(z+x)+(x+y)]\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x +y}\right)\geq(x+y+z)^2$

or $\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge q\frac{x+y+z}{2}\geq\frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}$

3. Since $abc=1\implies(abc)^2=1,$ both sides positives since the assumption $a,b,c\in\mathbb R^+$ is given, so the LHS is $\frac{a^{2}b^{2}c^{2}}{a^{3}(b+c)}+\frac{a^{2}b^{2 }c^{2}}{b^{3}(a+c)}+\frac{a^{2}b^{2}c^{2}}{c^{3}(a +b)}=\frac{(bc)^{2}}{a(b+c)}+\frac{(ac)^{2}}{b(a+c )}+\frac{(ab)^{2}}{c(a+b)}.$ Now apply Cauchy–Schwarz and get $\frac{(bc)^{2}}{a(b+c)}+\frac{(ac)^{2}}{b(a+c)}+\f rac{(ab)^{2}}{c(a+b)}\ge \frac{(ab+bc+ac)^{2}}{2(ab+bc+ac)}=\frac{ab+bc+ac} {2}.$ Hence since $ab+bc+ac\ge 3\sqrt[3]{(abc)^2}=3,$ we have our inequality.