a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0.
\] My way.
We use the usual substitution, $(a,b,c)=(x+y,x+z,y+z)$, $ x,y,z>0$, then the original inequality is equivalent to \[
x^3y+y^3z+z^3x\ge xyz(x+y+z).
\] Finally, the last inequality is easy enough to see by noting that \[
x^3y+y^3z+z^3x = xyz\left(\frac{x^2}{z}+\frac{y^2}{x}+\frac{z^2}{y}\right),
\] we are done.
We are learning how to solve ODE. Now for those who has learnt ODE in secondary school (took AL applied maths paper II), you must be able to do the following.
- Evaluate $\displaystyle u(x)=1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\cdots$, how do you link it with solving ODE? Of couse the substitution of cube root of unity will do in this question.
- Let $ f (x)$ be a differentiable function on $ [0, 1], f (0) = 0 $ and $f (1) = 1$, prove that $\displaystyle \int_0^1|f(x)-f'(x)|\,dx\ge \frac{1}{e}$. Once again, technique involved in solving ODE is needed. Can you recall it?
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