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.
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.
If $b=c=\frac12$ the LHS is $0$. So if you make them just a bit smaller than $\frac12$ what happens?
$b=0.5$ 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
Counterexample: $b=0.6, c=0.2$
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}