Given $abc=1$, prove that $${\frac{a}{a^2+2}}+ {\frac{b}{b^2+2}}+{\frac{c}{c^2+2}} \leqslant 1 $$
I tried substitution $a={\frac{1}{x}},b={\frac{1}{y}},c={\frac{1}{z}}$ but can't finish.
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I suspect that https://math.stackexchange.com/q/2090810/86846 will have some similar features. – abiessu Feb 11 '21 at 16:08
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Let $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$ we have to prove
$$\sum_{cyc}\frac{xy}{x^2+2y^2}\le 1$$ Now as $x^2+y^2\ge 2xy$ it suffices to prove
$$\sum_{cyc}\frac{x}{2x+y}\le 1 $$ $$\iff \sum_{cyc} \frac{y}{2x+y}\ge 1$$ which is true as
$$\sum_{cyc} \frac{y}{2x+y}= \sum_{cyc} \frac{y^2}{2xy+y^2}\ge \frac{{(x+y+z)}^2}{2xy+y^2+2zx+x^2+2yz+z^2}=1$$
Albus Dumbledore
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@Dr.Mathva if you have time please checkout this problem ,https://math.stackexchange.com/questions/4022005/proving-sum-cyc-fracaa22bcbc-ge-fracabc22 You may come up with an ingeniuos proof as you did before in another question of mine – Albus Dumbledore Feb 11 '21 at 17:50
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Hmm, I've been trying it for a while, but it seems to be strong. CS and Jensen have both failed... – Dr. Mathva Feb 11 '21 at 21:28
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How did you get that one sum is less than one which is equivalent to a different sum larger than one? – gournge Feb 26 '21 at 10:14
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Is this good ? $$\sum_{cyc} \frac{a}{a^2+2}=\sum_{cyc}\frac{a}{a(a+2bc)}= \sum_{cyc}\frac{1}{a+2bc} $$ using AM-GM : $$a+2bc \geq 2\sqrt{2}$$ so $$\sum_{cyc}\frac{1}{a+2bc} \le \frac{1}{2\sqrt{2}}\le 1$$ i dont know where but I’m sure that there is some mistake somewhere. – PNT Mar 01 '21 at 22:41
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