I'm new to proving inequalities. How does one prove this?

Question

If a, b, and c are non-negative real numbers and $a + b + c = 2 $, prove that $ 2 \ge a^2 b^2 + b^2 c^2 + c^2 a^2 $

Answer 1

It's actually $1\ge a^2 b^2 + b^2 c^2 + c^2 a^2$.

It is enough to show the homogenous symmetric inequality $$ (a+b+c)^4 - 16(a^2 b^2 + b^2 c^2 + c^2 a^2)\ge 0$$

Assume an ordering $a \le b \le c$. Write \begin{eqnarray} a &=&p \\ b&=& p+ q\\ c &=& p+q +r \end{eqnarray} with $p$, $q$, $r\ge 0$. Substitute $a$, $b$, $c$ in terms of $p$, $q$, $r$ in the expression above. We get a polynomial of degree $4$ in $p$, $q$, $r$ with all the coefficients positive. Since $p$, $q$, $r$ are also positive we conclude that the expression is positive.

Obs: We have equality if and only if two of the $a$, $b$, $c$ are equal and the third is $0$.

Answer 2

We'll prove that $$(a+b+c)^4\geq16(a^2b^2+a^2c^2+b^2c^2).$$ Indeed, by AM-GM we obtain: $$(a+b+c)^4=((a^2+b^2+c^2)+2(ab+ac+bc))^2\geq$$ $$\geq\left(2\sqrt{2(a^2+b^2+c^2)(ab+ac+bc)}\right)^2=8\sum_{cyc}(a^3b+ab^3+a^2bc)\geq$$ $$\geq8\sum_{cyc}(2a^2b^2+0)=16(a^2b^2+a^2c^2+b^2c^2).$$ Done!