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With $a + b + c = 3$ and $a, b, c>0$ prove these inequality:

1)$$\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$$

2)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+abc\geq 4$$

3)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+\frac{9}{4}abc\geq \frac{21}{4}$$

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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  Did Nov 12 '12 at 12:39
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Sorry about that, I am not English native language speaker and I just a secondary student so my English is not good; can u rewrite it for me ? Give me some suggestion ? –  Xeing Nov 12 '12 at 12:48
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Let me suggest that you start by adding the context and your thoughts on this question, then we will deal with the language. –  Did Nov 12 '12 at 12:53

2 Answers 2

up vote 6 down vote accepted

(Note: The notation $\sum$ refers to a cyclic sum here.)

For the first one: We can assume WLOG that $a\ge b\ge c$. We will prove a lemma: Let $x, y$ be positive reals with $xy\ge 1$. Then: $$\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge \frac{2}{1+xy}$$

Proof: Upon direct expanding, the inequality is equivalent to $(x-y)^2(xy-1)\ge 0$, which is indeed true.

Now applying the lemma with $x=\sqrt{\frac{ab}{c}}, y=\sqrt{\frac{ac}{b}}$, we have $xy=a\ge 1$ and thus: $$\sum\frac{a}{a+bc}=\sum\frac{1}{1+\frac{bc}{a}}\ge \frac{2}{1+a}+\frac{a}{a+bc}$$ Note that $bc\le\frac{(b+c)^2}{4}=\frac{(3-a)^2}{4}=\frac{a^2-6a+9}{4}$. Thus: $$\frac{2}{1+a}+\frac{a}{a+bc}\ge \frac{2}{1+a}+\frac{4a}{a^2-2a+9}$$ It suffices to show that the above is at least $\frac{3}{2}$. Indeed, direct calculation gives: $$2(1+a)(a^2-2a+9)(\frac{2}{1+a}+\frac{4a}{a^2-2a+9}-\frac{3}{2})=3(3-a)(a-1)^2\ge 0$$ because $a\le 3$.

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For the second and the third question,a generalized version is presented here:

Let $a,b,c>0$ and $a+b+c=3$,then

$$f(a,b,c)=(\sum\frac{ab}{c})+\lambda abc\geq3+\lambda$$

for $1\leq \lambda\leq9/4$,where the constant $9/4$ is optimal.

It is easy to see that 9/4 is optimal(a simple comparation between $f(1,1,1)$ and $f(2,1/2,1/2)$).

Proof:

Without loss of generality,we assume that $a\geq b\geq c$.Then $1\leq a<3$.

$$f(a,b,c)=a(\frac{b}{c}+\frac{c}{b})+\frac{bc}{a}+\lambda abc=\frac{a(b+c)^2}{bc}+(\lambda a+\frac{1}{a})bc-2a$$

Recall that $a+b+c=3$,then

$$f(a,b,c)=\frac{a(3-a)^2}{bc}+(\lambda a+\frac{1}{a})bc-2a$$

Given $p,q>0$,it is obvious that function $u(x)=px+q/x$ is monotone decreasing on interval $(0,\sqrt{q/p}]$.We take $(\lambda a+\frac{1}{a},a(3-a)^2)$ as $(p,q)$,then $f(a,b,c)$ is a monotone decreasing function(for $bc$) on the interval $(0,\sqrt{q/p}]$ with $a$ fixed.

AM-GM inequality suggests that $$bc\leq\frac{(b+c)^2}{4}=\frac{(3-a)^2}{4},$$ and it is natural to test whether $(b+c)^2/4\leq\sqrt{q/p}$,i.e., $$\frac{(3-a)^2}{4}\leq\frac{a(3-a)}{\sqrt{\lambda a^2+1}}$$

It suffices to show that $$\frac{(3-a)^2}{16}\leq\frac{a^2}{\lambda a^2+1},$$i.e.,

$$\lambda\frac{(3-a)^2}{16}+\frac{1}{\lambda a^2+1}\leq 1$$.

Because $1\leq a<3$,$LHS\leq\lambda/4+1/(\lambda+1)\leq 1$ for every $1\leq \lambda\leq 3$

Hence $f(a,b,c)$ achieve its minimum(for fixed $a$) when $bc=(3-a)^2/4$.

$$\min_{a fixed}f(a,b,c)=(\lambda a+\frac{1}{a})\frac{(3-a)^2}{4}+2a$$.

We just need to show that

$$(\lambda a+\frac{1}{a})\frac{(3-a)^2}{4}+2a\geq 3+\lambda$$

for every $1\leq \lambda\leq 9/4$,which is equivalent to show that

$$\lambda(a-1)^2(a^2-4a+\frac{9}{\lambda})\geq 0$$

for every $1\leq \lambda\leq 9/4$.

It suffices to prove that $(a^2-4a+\frac{9}{\lambda})\geq 0$ for for every $1\leq \lambda\leq 9/4$,and it is quite obvious.

Q.E.D.

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