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 –
- Let $x$,$y$,$z$ be three non-negative real numbers, prove that:
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
The first problem.
Your proof is beautiful!
My solution:
We need to prove that: $$\sum_{cyc}\frac{a^2+ac-2ab}{ab+1}\geq0$$ or $$\sum_{cyc}\frac{a^2+ac+2}{ab+1}\geq6.$$ Now, by C-S $$\sum_{cyc}\frac{a^2+ac+2}{ab+1}\geq\frac{\left(\sum\limits_{cyc}(a^2+ac+2)\right)^2}{\sum\limits_{cyc}(a^2+ac+2)(ab+1)}.$$ Thus, it's enough to prove that: $$\frac{\left(\sum\limits_{cyc}(a^2+ac+2)\right)^2}{\sum\limits_{cyc}(a^2+ac+2)(ab+1)}\geq6$$ or $$\sum_{cyc}(a^4-4a^3b+2a^3c+3a^2b^2-2a^2bc)+6\sum_{cyc}(a^2-ab)\geq0$$ and since $$6\sum_{cyc}(a^2-ab)=3\sum_{cyc}(a-b)^2\geq0,$$ it's enough to prove that $$\sum_{cyc}(a^4-4a^3b+2a^3c+3a^2b^2-2a^2bc)\geq0,$$ which is smooth.
The second problem.
By your work it's enough to prove that: $$\frac{3}{8}\sum_{cyc}(x+3)+\sum_{cyc}\frac{6}{x+3}\geq9,$$ which is true by C-S and AM-GM: $$\frac{3}{8}\sum_{cyc}(x+3)+\sum_{cyc}\frac{6}{x+3}=\frac{3}{8}\left(\sum_{cyc}(x+3)+16\sum_{cyc}\frac{1}{x+3}\right)\geq$$ $$\geq \frac{3}{8}\left(\sum_{cyc}(x+3)+\frac{144}{\sum\limits_{cyc}(x+3)}\right)\geq\frac{3}{8}\cdot2\cdot\sqrt{144}=9.$$