Proving 3(1−a+a2)(1−b+b2)(1−c+c2)≥1+abc+a2b2c2My task was to prove the question above over real variables.I thought that this minor inequality should help3(1−a+a2)(1−b+b2)≥2(1−ab+a2b2).which is true.By this inequality, the original inequality is converted to(1−ab)2(1−c)2+(ab−c)2+abc≥0This proves the inequality for abc≥0.I want to prove this Inequality for abc&lt;0. But I couldn't find a solution for abc&lt;0.Any extensions for abc&lt;0 are thankfully accepted.

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Proving 3(1−a+a2)(1−b+b2)(1−c+c2)≥1+abc+a2b2c2My task was to prove the question above over real variables.I thought that this minor inequality should help3(1−a+a2)(1−b+b2)≥2(1−ab+a2b2).which is true.By this inequality, the original inequality is converted to(1−ab)2(1−c)2+(ab−c)2+abc≥0This proves the inequality for abc≥0.I want to prove this Inequality for abc&amp;lt;0. But I couldn&#039;t find a solution for abc&amp;lt;0.Any extensions for abc&amp;lt;0 are thankfully accepted.

Avery Maxwell · Accepted Answer

Your first step leads to a wrong inequality because it does not save the case of the equality occurring:&amp;nbsp;a=b=c=1.After your first step it&#039;s enough to prove that:2(1−ab+a2b2)(1−c+c2)≥1+abc+a2b2c2,which is wrong for&amp;nbsp;a=b=c=1.The Vasc&#039;s solution.Since&amp;nbsp;2(a2−a+1)(b2−b+1)≥a2b2+1,it&#039;s enough to prove that:3(a2b2+1)(c2−c+1)≥2(a2b2c2+abc+1),which is a quadratic inequality of c.

membatas0v2v · Accepted Answer

Another way.It&#039;s enough to prove our inequality for non-negatives a, b and c.Now, since&amp;nbsp;3(a2−a+1)3−a6−a3−1=(a−1)4(2a2−a+2)≥0,by Holder we obtain:∏cyc(a2-a+1)≥∏cyca6+a3+133≥13(a2b2c2+abc+1).Now, let&amp;nbsp;a≤0,b≥0&amp;nbsp;and&amp;nbsp;c≥0.Thus, after replacing a on -a we need to prove that:3∑cyc(a2+a+1)(b2−b+1)(c2−c+1)≥a2b2c2−abc+1,which follows from the previous inequality:3∑cyc(a2+a+1)(b2−b+1)(c2−c+1)≥≥3∑cyc(a2−a+1)(b2−b+1)(c2−c+1)≥a2b2c2+abc+1≥a2b2c2−abc+1.

Proving \(\displaystyle{3}{\left({1}-{a}+{a}^{{2}}\right)}{\left({1}-{b}+{b}^{{2}}\right)}{\left({1}-{c}+{c}^{{2}}\right)}≥{1}+{a}{b}{c}+{a}^{{2}}{b}^{{2}}{c}^{{2}}\) My task was to prove the

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New Questions in Algebra I