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#1 h Oct 3, 2009, 4:06 AM • 2 Y

Y by Adventure10, Mango247

Given $a, b, c \ge 0$ and $k \ge 2$ . Prove that:
i) $\frac {ab(a + b)}{a^2 + b^2 + 2c^2} + \frac {bc(b + c)}{b^2 + c^2 + 2a^2} + \frac {ca(c + a)}{c^2 + a^2 + 2b^2} \le \frac {a + b + c}{2}$

ii) $\frac {a^k(b + c)}{b^2 + c^2} + \frac {b^k(c + a)}{c^2 + a^2} + \frac {c^k(a + b)}{a^2 + b^2} \ge a^{k - 1} + b^{k - 1} + c^{k - 1}$

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kuing

#2 h Oct 3, 2009, 5:56 AM • 2 Y

Y by Adventure10, Mango247

nguoivn wrote:

Given $a, b, c \ge 0$ and $k \ge 2$ . Prove that:
i) $\frac {ab(a + b)}{a^2 + b^2 + 2c^2} + \frac {bc(b + c)}{b^2 + c^2 + 2a^2} + \frac {ca(c + a)}{c^2 + a^2 + 2b^2} \le \frac {a + b + c}{2}$

ii) $\frac {a^k(b + c)}{b^2 + c^2} + \frac {b^k(c + a)}{c^2 + a^2} + \frac {c^k(a + b)}{a^2 + b^2} \ge a^{k - 1} + b^{k - 1} + c^{k - 1}$

second one:

$\frac {a^k(b + c)}{b^2 + c^2} + \frac {b^k(c + a)}{c^2 + a^2} + \frac {c^k(a + b)}{a^2 + b^2} \ge a^{k - 1} + b^{k - 1} + c^{k - 1}$

$\Longleftrightarrow \sum {\frac {{ab\left( {\left( {a - b} \right)\left( {a^k - b^k } \right) + c^2 \left( {a - b} \right)\left( {a^{k - 2} - b^{k - 2} } \right)} \right)}}{{\left( {b^2 + c^2 } \right)\left( {c^2 + a^2 } \right)}}} \ge 0.$
for $k\ge2$ , whether $a\le b$ or $a\ge b$ , we also have ${\left( {a - b} \right)\left( {a^k - b^k } \right)}\ge 0$ and ${\left( {a - b} \right)\left( {a^{k - 2} - b^{k - 2} } \right)}\ge 0$ , so the inequality holds...

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#3 h Oct 3, 2009, 7:40 AM • 2 Y

Y by Adventure10, Mango247

nguoivn wrote:

Given $a, b, c \ge 0$ and $k \ge 2$ . Prove that:
i) $\frac {ab(a + b)}{a^2 + b^2 + 2c^2} + \frac {bc(b + c)}{b^2 + c^2 + 2a^2} + \frac {ca(c + a)}{c^2 + a^2 + 2b^2} \le \frac {a + b + c}{2}$

This is not true, nguoivn. A counterexample is $(a,b,c)=(0.5,1,1).$

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#4 h Oct 3, 2009, 10:31 AM • 2 Y

Y by Adventure10, Mango247

can_hang2007 wrote:

nguoivn wrote:

Given $a, b, c \ge 0$ and $k \ge 2$ . Prove that:
i) $\frac {ab(a + b)}{a^2 + b^2 + 2c^2} + \frac {bc(b + c)}{b^2 + c^2 + 2a^2} + \frac {ca(c + a)}{c^2 + a^2 + 2b^2} \le \frac {a + b + c}{2}$

This is not true, nguoivn. A counterexample is $(a,b,c) = (0.5,1,1).$

Sorry, I think the easier is trues

:
$\frac {ab(a + b)}{2a^2 + 2b^2 + 5c^2} + \frac {bc(b + c)}{2b^2 + 2c^2 + 5a^2} + \frac {ca(c + a)}{2c^2 + 2a^2 + 5b^2} \le \frac {a + b + c}{4}$

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