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I am trying to prove (or disprove) this inequality for more than one hour without success.

$$[1-(b+c)]^2+[1-2c]^2\ge 4bc$$ where $b,c>0$ and satisfies $b+c<1$.

Frustratingly, I failed to find a counterexample.

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If $b=c=\frac12$ the LHS is $0$. So if you make them just a bit smaller than $\frac12$ what happens?

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  • $\begingroup$ But we need $b + c < 1$. $\endgroup$ – Mikko Korhonen Dec 21 '11 at 15:33
  • $\begingroup$ @m.k.: Your comment and my edit crossed in etherspace $\endgroup$ – Ross Millikan Dec 21 '11 at 15:35
  • $\begingroup$ My bad, I missed it. Thanks! $\endgroup$ – Tapu Dec 21 '11 at 15:44
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Counterexample: $b=0.6, c=0.2$

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$b=0.5$ and $c=0.3$ gives $0.2^2 + 0.4^2 = 0.2 < 0.6 = 4\times 0.5\times 0.3$ and there are many more counter-examples

Indeed if you choose a value for $c$ in the range $\frac{1}{2}-\sqrt{\frac{1}{8}} \lt c \lt \frac{1}{2}+\sqrt{\frac{1}{8}}$ and then choose a value for $b$ in the range $1+c-\sqrt{-4\,{c}^{2}+8\,c-1} \lt b \lt 1-c$ you should find the direction of the inequality seems to be reversed

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Taking $b=c=a\in (1-\frac{\sqrt{2}}{2},\frac{1}{2})\subset(1-\frac{\sqrt{2}}{2},1+\frac{\sqrt{2}}{2})$. We are easily to verify that $b,c>0$ and $b+c<1$. Moreover, in this case we also have \begin{eqnarray} \text{LHS}&=&(1-2a)^2+(1-2a)^2\\ &=&(4a^2-8a+2)+4a^2\\ &=&4(a-(1+\frac{\sqrt{2}}{2}))(a-(1-\frac{\sqrt{2}}{2}))+4a^2\\ &<&\text{RHS}. \end{eqnarray}

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  • $\begingroup$ As the above arguments, there are countless ways of choosing $(b,c)$ such that the given inequalities are not satisfied. $\endgroup$ – blindman Jul 27 '12 at 1:55

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