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Inequality 5(Vasile Cirtoaje)


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.


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.


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