# Proving the inequality $4\ge a^2b+b^2c+c^2a+abc$

So, a,b,c are non-negative real numbers for which holds that $a+b+c=3$. Prove the following inequality: $$4\ge a^2b+b^2c+c^2a+abc$$

For now I have only tried to write the inequality as $$4\left(\frac{a+b+c}3\right)^3\ge a^2b+b^2c+c^2a+abc$$ but I don't know what to do after that...

So we need to show $$4(a+b+c)^3 \ge 27(a^2b+b^2c+c^2a+abc)$$

One way is to use the cyclic symmetry and WLOG assume $a$ to be the min of $a, b, c$. Then we can write $b = a+x, c = a+y$, where $x, y \ge 0$. Now the inequality reduces to $$9a(x^2-xy+y^2)+(x-2y)^2(4x+y) \ge 0$$ which is obvious. Also from the above, we get that equality is possible iff $x=y=0$ or when $a=0, x=2y$, i.e. when $(a, b, c)=(1, 1, 1)$ or is a permutation of $(0, 2, 1)$.

• Today in school I was thinking about this problem again and I came with the same solution, just with AM-GM. Nice work there :) Nov 19, 2014 at 14:30

Let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.

Hence, by Rearrangement and AM-GM we obtain:

$a^2b+b^2c+c^2a+abc=a\cdot ab+b\cdot bc+c\cdot ca+abc\leq x\cdot xy+y\cdot xz+z\cdot yz+xyz=$

$=y(x+z)^2=4y\left(\frac{x+z}{2}\right)^2\leq4\left(\frac{y+\frac{x+z}{2}+\frac{x+z}{2}}{3}\right)^3=4$.