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a+b+c=3 ine
jokehim   4
N Mar 21, 2025 by lbh_qys
Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c=3.$ Prove that $$\color{black}{\frac{a\left(b+c\right)}{bc+3}+\frac{b\left(c+a\right)}{ca+3}+\frac{c\left(a+b\right)}{ab+3}\le \frac{3}{2}.}$$Proposed by Phan Ngoc Chau
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jokehim
Mar 18, 2025
lbh_qys
Mar 21, 2025
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a+b+c=3 ine
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jokehim
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jokehim
#1 h  Mar 18, 2025, 9:48 AM
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Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c=3.$ Prove that $$\color{black}{\frac{a\left(b+c\right)}{bc+3}+\frac{b\left(c+a\right)}{ca+3}+\frac{c\left(a+b\right)}{ab+3}\le \frac{3}{2}.}$$Proposed by Phan Ngoc Chau
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Cool12345678
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Cool12345678
#2 h  Mar 18, 2025, 3:04 PM
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Since $(2,0,0)$ majorizes $(1,1,0)$ and given the condition $a+b+c=3$, by Muirhead, we rewrite $$\sum_{cyc} \frac{a\left(b+c\right)}{bc+3}\leq\sum_{cyc} \frac{a\left(3-a\right)}{a^2+3}$$
Now, let $f(x)=\frac{x\left(3-x\right)}{x^2+3}$. Then, setting $y=\frac{x\left(3-x\right)}{x^2+3}$, we get $$ x^2y + 3y = 3x-x^2 \implies x^2(y + 1) + 3y-3x = 0.$$Using the discriminant of a quadratic, we get $$9-4(y+1)(3y)\geq0 \implies y\leq \frac{1}{2}$$Therefore, the maximum value of $f(x)$ is $\frac{1}{2}$ so for all $a,b,c$ we have $$\frac{a\left(3-a\right)}{a^2+3}\leq \frac{1}{2}$$Hence, $$\sum_{cyc} \frac{a\left(3-a\right)}{a^2+3}\leq \frac{3}{2}$$as desired.
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no_room_for_error
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no_room_for_error
#3 h  Mar 18, 2025, 7:20 PM
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Cool12345678 wrote:
Since $(2,0,0)$ majorizes $(1,1,0)$ and given the condition $a+b+c=3$, by Muirhead, we rewrite $$\sum_{cyc} \frac{a\left(b+c\right)}{bc+3}\leq\sum_{cyc} \frac{a\left(3-a\right)}{a^2+3}$$

You can try $(a,b,c)=(0,1,2).$
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jokehim
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#4 h  Mar 21, 2025, 5:45 AM
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Any ideas?
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lbh_qys
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lbh_qys
#5 h  Mar 21, 2025, 8:26 AM
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Let
\[ q = ab+bc+ca,\quad r = abc. \]Adding 1 to each fraction, we obtain
\[ \sum \frac{q+3}{bc+3} \le \frac{9}{2}, \]that is,
\[ \sum \frac{1}{bc+3} \le \frac{9}{2q+6}. \]Upon combining the terms on the left-hand side, the numerator is
\[ \sum (ab+3)(ac+3)=3r+6q+27, \]and the denominator is
\[ \prod (ab+3)=r^2+9r+9q+27. \]Hence, it is equivalent to proving that
\[ \frac{3r+6q+27}{r^2+9r+9q+27}\le \frac{9}{2q+6}. \]This simplifies to
\[ 4q^2+3q\le 3r^2+(21-2q)r+27. \]In fact, since \(q\le 3\) and by Schur’s inequality \(3r\ge 4q-9\), it follows that
\[ 3r^2+(21-2q)r+27\ge (4q-9)r+(21-2q)r+27=(2q+12)r+27\ge 3(q+3)r+27\ge (q+3)(4q-9)+27=4q^2+3q. \]Thus, the original inequality holds, with equality when \((a,b,c)=(1,1,1)\) or \((a,b,c)=(3/2,3/2,0)\) and their permutations.
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