Prove that if $a+b+c=1$ then $\frac{ab}{c+1}+\frac{ac}{b+1}+\frac{bc}{a+1} \le \frac{1}4$ $a,b,c$ are positive real numbers

Question

I simplified the inequality , but I didn't get anything that may seem useful.

I got

$$^\sum_{cyc} 4a^2b+8ab+4ab^2\le abc+ab+ac+ab+a+b+c+1$$

Are there any other conditions for $a$ $b$ and $c$? If not $a = b = 0$ and $c=1$ would be a counter example. — Sqyuli, Nov 11 '18 at 10:42

score 1 · Accepted Answer · answered Nov 11 '18 at 10:52

1

For positive variables the reversed inequality is true.

Indeed, by C-S we obtain: $$\sum_{cyc}\frac{ab}{1+c}=\sum_{cyc}\frac{ab}{2c+a+b}\leq\frac{1}{4}\sum_{cyc}ab\left(\frac{1^2}{c+a}+\frac{1^2}{c+b}\right)=$$ $$=\frac{1}{4}\sum_{cyc}\left(\frac{ab}{c+a}+\frac{bc}{a+c}\right)=\frac{1}{4}\sum_{cyc}b=\frac{1}{4}.$$

answered Nov 11 '18 at 10:52

Michael Rozenberg

194,933

1

Indeed: https://math.stackexchange.com/a/2225376/42969. – Martin R Nov 11 '18 at 17:16

Prove that if $a+b+c=1$ then $\frac{ab}{c+1}+\frac{ac}{b+1}+\frac{bc}{a+1} \le \frac{1}4$ $a,b,c$ are positive real numbers

1 Answers1