Leta; b; c be positive reals such that abc = 1. Prove that
{1/sqrt(b + a/2 + 1/2)} + {1/sqrt(c + b/2 + 1/2)} + {1/sqrt(a + c/2 + 1/2)} >= sqrt(2)
Advanced Algebra?
I only studied retarded/subdued/garden-variety Algebra. So I'll attack your question here with my crude tools, and then I'll leave it hanging for others who were/are into Advanced Algebra.
{1/sqrt(b + a/2 + 1/2)} + {1/sqrt(c + b/2 + 1/2)} + {1/sqrt(a + c/2 + 1/2)} >= sqrt(2)
{1/sqrt[(2b+a+1)/2]} + {1/sqrt[(2c+b+1)/2]} + {1/sqrt[(2a+c+1)/2]} >= sqrt(2)
{sqrt(2)/sqrt(2b+a+1)} + {sqrt(2)/sqrt(2c+b+1)} + {sqrt(2)/sqrt(2a+c+1)} >= sqrt(2)
{1/sqrt(2b+a+1)} + {1/sqrt(2c+b+1)} + {1/sqrt(2a+c+1)} >= 1
I'm up to there only.
Maybe I'd expand that by eliminating all the radical signs, but I'm not yet too crazy.
{1/sqrt(2b+a+1)} + {1/sqrt(2c+b+1)} + {1/sqrt(2a+c+1)} >= abc ?
I don't know that too.
well for starters I advise that you don't begin with AM-GM. instead use AM-HM
which says that for positive $\displaystyle \alpha$, $\displaystyle \beta$, $\displaystyle \gamma$ that $\displaystyle \frac{\alpha+\beta+\gamma}{3}\geq \frac{3}{\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1} {\gamma}}$
so using this we that the expression on the left (with out the square roots) is greater than or equal to
$\displaystyle \frac{9}{a+b+c+\frac{\alpha}{2}+\frac{\beta}{2}+\f rac{\gamma}{2}+\frac{3}{2}}$ which is less than
$\displaystyle \frac{9}{a+b+c+\frac{\alpha}{2}+\frac{\beta}{2}+\f rac{\gamma}{2}}$
NOW from AM-GM we have that this is less than $\displaystyle \frac{9}{3+\frac{3}{2}}=2$
now that you know this you can do a little modification to get $\displaystyle \sqrt{2}$