How to prove common cubic inequalities involving a, b, and c

Expert intro: These inequalities look different, but the same trick keeps showing up: factor the difference, then notice a square or a product of nonnegative terms. Once you know what to look for, the pattern becomes much easier to spot.

Detailed walkthrough

Here are clean ways to prove each one.

1) Prove $a^3+b^3\ge a^2b+ab^2$

Move everything to one side:

$a^3+b^3-a^2b-ab^2.$

Factor by grouping:

$a^2(a-b)-b^2(a-b)=(a-b)(a^2-b^2)=(a-b)^2(a+b).$

So the inequality is equivalent to

$(a-b)^2(a+b)\ge 0.$

This is true whenever $a+b\ge 0$. In particular, for nonnegative $a,b$, it holds.

2) Prove $a^3+b^3+c^3\ge a^2b+b^2c+c^2a$

A useful identity is

$(a^3+b^3+c^3- a^2b-b^2c-c^2a)=\frac12\sum_{cyc}(a-b)^2(a+b-c).$

So the inequality is true under conditions such as the triangle-type constraints where each $(a+b-c)\ge 0$. If no extra assumptions are given, it is not true in full generality.

So this one needs a condition on $a,b,c$; otherwise a counterexample may exist.

3) Prove $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2$

Assume $a,b,c>0$. By AM-GM,

$\frac{a^3}{b}+b \ge 2a,$

but a better route is to use AM-GM in cyclic form:

\quad \frac{b^3}{c}+\frac{c}{1}\ge 2b, \quad \frac{c^3}{a}+\frac{a}{1}\ge 2c.$$ That is not yet enough directly, so a standard substitution helps: let $$x=\frac{a}{\sqrt{b}},\quad y=\frac{b}{\sqrt{c}},\quad z=\frac{c}{\sqrt{a}}.$$ Then the left side becomes $x^4+y^4+z^4$-type terms after normalization, and by AM-GM / power mean, each term dominates the corresponding square term. A more direct proof uses the uvw or Cauchy framework under positivity assumptions. If you want a sharper elementary proof, I can rewrite this one step by step with the exact assumptions your class expects. ### 💡 Pitfall guide The second and third inequalities are not automatically true without conditions. A lot of students try to prove them as if they were universal identities, but the sign of the expression can change if the variables are unrestricted or if some denominators are negative. Always check the hidden assumptions first. ### 🔄 Real-world variant If the problem assumes $a,b,c\ge 0$, then the first inequality is immediate from factoring, while the third can often be attacked by AM-GM or Cauchy. If a condition like $a,b,c$ form sides of a triangle is included, the cyclic cubic inequality in part 2 becomes much more manageable. ### 🔍 Related terms cyclic inequality, AM-GM inequality, factorization