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# Inequality 5(Vasile Cirtoaje)

Problem:

If $\displaystyle a,b,c$ are non-negative numbers, prove that

$\displaystyle 3(a^2-a+1)(b^2-b+1)(c^2-c+1)\geq 1+abc+a^2b^2c^2$.

Solution:

From the identity

$\displaystyle 2(a^2-a+1)(b^2-b+1)=1+a^2b^2+(a-b)^{2}+(1-a)^{2}(1-b)^{2}$

follows the inequality

$\displaystyle 2(a^2-a+1)(b^2-b+1)\geq 1+a^2b^2$.

Thus, we only need to prove that

$\displaystyle 3(1+a^2b^2)(c^2-c+1)\geq 2(1+abc+a^2b^2c^2)$

which is equivalent to the quadratic in $\displaystyle c$ equation

$\displaystyle (3+a^2b^2)c^2-(3+2ab+3a^2b^2)c+1+3a^2b^2\geq 0$,

which is true since the discriminant $\displaystyle D$ is equal to $\displaystyle D=-3(1-ab)^{4}\leq 0$.

Equality occurs for $\displaystyle a=b=c=1$, Q.E.D.