These are two inequalities from my assignments. I don't know if it is too difficult or I am not so good at inequalities but please help me with full answers.
- Let $a$,$b$,$c$ be three real positive numbers, prove that
So first, to avoid typing too much character, I may write the LHS of the first problem in the form of $\sum\frac{a(a−2b+c)}{a×b+1}$, which I transformed to $\sum\frac{a^2+1}{a×b+1} + \sum\frac{a×c+1}{a×b+1}−6.$
It is easy to prove using a.m-g.m inequality that $\sum\frac{a×c+1}{a×b+1}\geq3.$
It is also clear that $\sum\frac{a^2+1}{a×b+1}\geq3$ by Cauchy-Schwartz Inequality. Q.E.D –
The second problem is a little more complicated, I think so I haven't finished it yet, but still, this is what I have worked on:
$\sum\frac{x}{\sqrt{x^2+3}}\leq\frac{3\sqrt3}{8}\sqrt{x^2+y^2+z^2}+\frac{3}{8}.$
According to the Cauchy-Schwartz inequality, $\sqrt{(x^2+3)(1+3)}\geq x+3$,
$\sqrt{3(x^2+y^2+z^2)}\geq x+y+z$; further transformations yield,
$\frac{3}{8}\sum(x+3)+\sum\frac{6}{x+3}\geq9$. This seems so obvious but how do I finish it?
POST SCRIPT I really look forward to you posting your full answers rather than comment on this post. Please help with this