# How to prove $\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$

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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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
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

(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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