\frac{a}{b^4+c^4+a}

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\frac{a}{b^4+c^4+a}

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#1 h Sep 21, 2010, 3:01 PM • 2 Y

Y by Adventure10, Mango247

Let $a, b, c$ be positive real numbers such that $abc = 1$ . Prove that

$\frac{a}{b^4+c^4+a}+\frac{b}{c^4+a^4+b}+\frac{c}{a^4+b^4+c}\leq 1$ .

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truongtansang89

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truongtansang89

#2 h Sep 21, 2010, 3:29 PM • 2 Y

Y by Adventure10, Mango247

Winner2010 wrote:

Let $a, b, c$ be positive real numbers such that $abc = 1$ . Prove that

$\frac{a}{b^4+c^4+a}+\frac{b}{c^4+a^4+b}+\frac{c}{a^4+b^4+c}\leq 1$ .

Using $b^4 + c^4 \ge bc(b^2+c^2)$ then:
$\sum{\frac{a}{b^4+c^4+a}} \le \sum{\frac{a}{bc(b^2+c^2)+a}} \le \sum\frac{a^2}{a^2+b^2+c^2} = 1$

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arqady

#3 h Sep 21, 2010, 8:46 PM • 1 Y

Y by Adventure10

Winner2010 wrote:

Let $a, b, c$ be positive real numbers such that $abc = 1$ . Prove that

$\frac{a}{b^4+c^4+a}+\frac{b}{c^4+a^4+b}+\frac{c}{a^4+b^4+c}\leq 1$ .

Generalization:
Let $a$ , $b$ and $c$ are positive numbers such that $abc = 1$ and $\alpha\geq1$ . Prove that

$\frac{a}{b^{\alpha}+c^{\alpha}+a}+\frac{b}{c^{\alpha}+a^{\alpha}+b}+\frac{c}{a^{\alpha}+b^{\alpha}+c}\leq 1$

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Obel1x

#4 h Sep 21, 2010, 11:30 PM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

Let $a$ , $b$ and $c$ are positive numbers such that $abc = 1$ and $\alpha\geq1$ .

Is $\alpha \ge 1$ a natural number or any real number ?

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#5 h Sep 22, 2010, 12:44 AM • 1 Y

Y by Adventure10

${\alpha}$ , or $k$ , is congruent 1 mod 3, then it's solved since we can turn $k=3r-2$ and have $(b^{r-1}c^{r-1})({b^r}+{c^r})$ , but if $k$ doesn't satisfy, we will get fractional exponents; however that souldn't be too much of a problem with substitution.

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arqady

#6 h Sep 22, 2010, 5:38 AM • 2 Y

Y by Adventure10, Mango247

Obel1x wrote:

arqady wrote:

Let $a$ , $b$ and $c$ are positive numbers such that $abc = 1$ and $\alpha\geq1$ .

Is $\alpha \ge 1$ a natural number or any real number ?

$\alpha$ is a real number.

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Obel1x

#7 h Sep 22, 2010, 7:28 AM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

Obel1x wrote:

arqady wrote:

Let $a$ , $b$ and $c$ are positive numbers such that $abc = 1$ and $\alpha\geq1$ .

Is $\alpha \ge 1$ a natural number or any real number ?

$\alpha$ is a real number.

Can you show the solution of the problem ?

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arqady

#8 h Sep 22, 2010, 7:46 AM • 2 Y

Y by Adventure10, Mango247

Obel1x wrote:

Can you show the solution of the problem ?

Maybe somebody likes to prove it. Try! It's not hard.

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kuing

#9 h Sep 22, 2010, 7:48 AM • 2 Y

Y by Adventure10, Mango247

same way

$\sum {\frac{a}{{b^\alpha + c^\alpha + a}}} \le \sum {\frac{a}{{b^{\frac{{\alpha - 1}}{3}} c^{\frac{{\alpha - 1}}{3}} \left( {b^{\frac{{\alpha + 2}}{3}} + c^{\frac{{\alpha + 2}}{3}} } \right) + a}}}$ $= \sum {\frac{{a^{\frac{{\alpha + 2}}{3}} }}{{b^{\frac{{\alpha + 2}}{3}} + c^{\frac{{\alpha + 2}}{3}} + a^{\frac{{\alpha + 2}}{3}} }}} = 1$

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