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

Marcinek665
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    Welcome to math.SE: since you are new, we 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. –  Jan 18 '13 at 18:33
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    Do you mean to say $a,b,c$ are positive numbers? – Maesumi Jan 18 '13 at 18:46
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    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
  • Well if $a,b,c$ are positive integers and add up to 3 then $a=b=c=1$. And then $3=3$. –  Jan 18 '13 at 18:42
  • My bad. I mean, of course, positive real, not integer. I've just edited my post. – Marcinek665 Feb 02 '13 at 20:47

3 Answers3

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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 $(1') + (2') + (3') + (4')$.

  • 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. –  Jan 18 '13 at 20:38
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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.

daniel
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By Holder $$\sum_{cyc}\frac{a+b+ab}{c^2+a^3+b^4}=\sum_{cyc}\frac{9(c^2+a+1)(a+b+ab)}{(c^2+a+1)(c^2+a^3+b^4)(1+1+1)(1+1+1)}\leq$$ $$\leq\frac{9\sum\limits_{cyc}(c^2+a+1)(a+b+ab)}{(a+b+c)^4}=\frac{1}{9}\sum_{cyc}(a^2b+a^2c+a^2bc+a^2+ab+a^2b+2+ab)=$$ $$=\frac{1}{9}\sum_{cyc}(2a^2b+a^2c+a^2bc+a^2+2ab+2)=\frac{1}{9}\sum_{cyc}(2a^2b+a^2c+abc+5)=$$ $$=\frac{2(a^2b+b^2c+c^2a+abc)+(a^2c+b^2a+c^2b+abc)+15}{9}\leq\frac{2\cdot4+4+15}{9}=3.$$ We used the following lemma.

Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c=3$. Prove that: $$a^2b+b^2c+c^2a+abc\leq4$$

A proof of the lemma.

Let $\{a,b,c\}=\{x,y,z\}$ such that $x\geq y\geq z$.

Hence, by Rearrangement and AM-GM we obtain:

$$a^2b+b^2c+c^2a+abc=a\cdot ab+b\cdot bc+c\cdot ca+xyz\leq x\cdot xy+y\cdot xz+z\cdot yz+xyz=$$ $$=y(x+z)^2=4y\left(\frac{x+z}{2}\right)^2\leq4\left(\frac{y+\frac{x+z}{2}+\frac{x+z}{2}}{3}\right)^3=4$$ Done!

Michael Rozenberg
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