<s>[INST] Proving that $\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\ge\frac32$ using derivatives 
Let $a,b,c\in\mathbb{R}^+$ and $abc=1$. Prove that
  $$\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\ge\frac32$$

This isn't hard problem. I have already solved it in following way:
Let $x=\frac1a,y=\frac1b,z=\frac1c$, then $xyz=1$. Now, it is enought to prove that
$$L\equiv\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge\frac32$$
Now using Cauchy-Schwarz inequality on numbers $a_1=\sqrt{y+z},a_2=\sqrt{z+x},a_3=\sqrt{x+y},b_1=\frac{x}{a_1},b_2=\frac{y}{a_2},b_3=\frac{z}{a_3}$ I got
$$(x+y+z)^2\le((x+y)+(y+z)+(z+x))\cdot L$$
From this
$$L\ge\frac{x+y+z}2\ge\frac32\sqrt[3]{xyz}=\frac32$$
Then I tried to prove it using derivatives. Let $x=a,y=b$ and
$$f(x,y)=\frac1{x^3\left({y+\frac1{xy}}\right)}+\frac1{y^3\left({x+\frac1{xy}}\right)}+\frac1{\left({\frac1{xy}}\right)^3(x+y)}$$
So, I need to find minimum value of this function. It will be true when
$$\frac{df}{dx}=0\land\frac{df}{dy}=0$$
After simplifying $\frac{df}{dx}=0$ I got
$$\frac{-y(3xy^2+2)}{x^3\left({xy^2+1}\right)^2}+\frac{1-x^2y}{y^2\left({x^2y+1}\right)^2}+\frac{x^2y^3(2x+3y)}{\left({x+y}\right)^2}=0$$
Is there any easy way to write $x$ in term of $y$ from this equation?
 [/INST] You can use this way to do. Your inequality $L\ge \frac32$ is equivalent to
$$ 2[x^2(x+y)(x+z)+y^2(x+y)(y+z)+z^2(x+z)(y+z)]\ge 3(x+y)(x+z)(y+z). $$
 Let
$$ F(x,y,z)=2[x^2(x+y)(x+z)+y^2(x+y)(y+z)+z^2(x+z)(y+z)]-3(x+y)(x+z)(y+z)-\lambda(xyz-1). $$
Then set
$$ \frac{\partial F}{\partial x}=0, \frac{\partial F}{\partial y}=0,\frac{\partial F}{\partial z}=0, \frac{\partial F}{\partial \lambda}=0. $$
Easy calculation shows that, $\frac{\partial F}{\partial x}=\frac{\partial F}{\partial y}$ gives $(x-y)[3(x+y)+\lambda z]=0.$
So $x=y$. Similarly $x=y=z$. But $xyz=1$ and hence $x=y=z=1$. So $f(x,y,z)$ reaches its minimum $0$ when $x=y=z=1$ or $f(x,y,z)\ge0$. Thus $L\ge\frac32$.
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