# Proving an inequality for positive numbers $a, b, c$

Let be $a,b,c$ positive numbers such that $a+b+c=3$. Prove that $$\frac{b+c+bc}{a^2+b^3+c^4}+\frac{c+a+ca}{b^2+c^3+a^4}+\frac{a+b+ab}{c^2+a^3+b^4} \le 3$$

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Well if $a,b,c$ are positive integers and add up to 3 then $a=b=c=1$. And then $3=3$. –  user58798 Jan 18 '13 at 18:42
Do you mean to say $a,b,c$ are positive numbers? –  Maesumi Jan 18 '13 at 18:46
The fact that the inequality has mixed degrees (not homogeneous) makes it interesting and rare, even though it is hard to see how it relates to any other topic. –  Maesumi Jan 18 '13 at 19:36
My bad. I mean, of course, positive real, not integer. I've just edited my post. –  Marcinek665 Feb 2 '13 at 20:47

As a habit I write x,y,z for a,b,c. Letting $z = 3 - (x+y)$ and plotting the function in Mathematica reveals a local maximum on $[\{0,3\},\{0,3\}]$ which does appear to be very close to $x= y = 1,$ and the maximum attained is of course 3.

The partial derivative of the l.h.s. of the OP with respect to y letting x = 1 is zero at y = 1. The partial derivative of the l.h.s. with respect to x at y = 1 is zero at x = 1.

On the domain of interest, it does then appear that $a = b = c= 1$ give a maximum of 3 and the objective function is less than three for $(x,y)\neq (1,1)$ and so the inequality is true.

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Using Cauchy Schwarz inequality: $(a^2+b^3+c^4)(a^2+b+1) \ge (a^2+b^2+c^2)^2$. Do this for each denominator, then it suffices to show that $$\sum (a^2+b+1)(b+c+bc) \leq 3(a^2+b^2+c^2)^2$$ where the sum is cyclic. Expand the left hand side, we get $$\begin{eqnarray} &\sum a^2b + b^2 + b + a^2c + bc + c + a^2bc + b^2c + bc \leq 3(a^2+b^2+c^2)^2 \\ \Leftrightarrow & 2 \sum a^2b + \sum a^2 c + \sum b^2 + \sum b + \sum c + 2\sum bc + \sum a^2bc \leq 3(a^2+b^2+c^2)^2 \\ \Leftrightarrow & \sum (a^2b+a^2c) + \sum a^2b + \sum a^2 + 3 + 3 + 2\sum ab + 3abc \leq 3(a^2+b^2+c^2)^2 \cdots (*) \end{eqnarray}$$ using $a+b+c = 3$. Now note the identity $$\sum(a^2b+a^2c) +3abc = a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+3abc = (ab+bc+ca)(a+b+c)$$ which is equal to $3(ab+bc+ca)$ for this problem, so $$(*)\Leftrightarrow \sum a^2 + 5 \sum ab + 6 + \sum a^2b \leq 3(a^2+b^2+c^2)^2$$ to be proved now.

Note the following inequalities: $$\begin{eqnarray}(1) & \hspace{5mm}(a^2+b^2+c^2)^2 +3 \ge 4 \sum a^2b \\ (2)& \hspace{5mm}(a^2+b^2+c^2) \ge 3 \\ (3)& \hspace{5mm}(a^2+b^2+c^2) \ge ab+bc+ca \end{eqnarray}$$ Here (1) follows from applying AM-GM to $a^4+a^2b+a^2b+1 \ge 4a^2b$ and sum up cyclicly. The other two are routine applications of Cauchy-Schwarz/AM-GM.

These implies $$\begin{eqnarray}(1') &\hspace{5mm} \frac{(a^2+b^2+c^2)^2}{3} \ge \frac{1}{4} (a^2+b^2+c^2)^2 + \frac{3}{4} \ge \sum a^2b \\ (2')&\hspace{5mm} \frac{2}{3}(a^2+b^2+c^2)^2 \ge 6 \\ (3')&\hspace{5mm} \frac{5}{3}(a^2+b^2+c^2)^2 \ge 5(a^2+b^2+c^2) \ge 5\sum ab \\ (4')&\hspace{5mm} \frac{1}{3} (a^2+b^2+c^2)^2 \ge \sum a^2 \end{eqnarray}$$ The inequality then follows from $(1') + (2') + (3') + (4')$.

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Is your Proof of Reduction to explain how CS leads the the first "suffices to show"? I do not really understand what you're trying to do. –  Calvin Lin Jan 18 '13 at 20:36
@CalvinLin, yes. Maybe I should remove the words reduction to avoid confusion. –  user27126 Jan 18 '13 at 20:38