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a+b+c=3 ine
jokehim   2
N Tuesday at 7:20 PM by no_room_for_error
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
no_room_for_error
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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  Tuesday at 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  Tuesday at 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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