A new symetric inequality - minhhoang

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A new symetric inequality - minhhoang

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#1 h Sep 18, 2010, 5:15 PM • 4 Y

Y by thangle, luofangxiang, Adventure10, Mango247

Let a,b,c be positive real numbers. Prove that:

$\frac{1}{{{a^2} + ab + {b^2}}} + \frac{1}{{{b^2} + bc + {c^2}}} + \frac{1}{{{c^2} + ca + {a^2}}} \le \frac{{2({k^2} + k + 1)}}{{{a^2} + {b^2} + {c^2} + ab + bc + ca}}$

With $k = \frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ca}}$

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#2 h Sep 19, 2010, 2:09 AM • 2 Y

Y by Adventure10, Mango247

minhhoang wrote:

Let a,b,c be positive real numbers. Prove that:

$\frac{1}{{{a^2} + ab + {b^2}}} + \frac{1}{{{b^2} + bc + {c^2}}} + \frac{1}{{{c^2} + ca + {a^2}}} \le \frac{{2({k^2} + k + 1)}}{{{a^2} + {b^2} + {c^2} + ab + bc + ca}}$

With $k = \frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ca}}$

We have: : $x^2+xy+y^2\ge \frac{3}{4}(x+y)^2$

$\Rightarrow LS\le \frac{4}{3}\sum\frac{1}{(a+b)^2}$

We need to prove:

$\frac{4}{3}\sum\frac{1}{(a+b)^2}\le ( \frac{{2({k^2} + k + 1)}}{{{a^2} + {b^2} + {c^2} + ab + bc + ca}}$ (1)

Suppose that: $a+b+c=1$

Put $p=a+b+c; q=ab+bc=ca; r=abc\Rightarrow p=1; r\le \frac{p}{9}$

By a number of simple transformation we have:

$(1) \Leftrightarrow \frac{(1-q)[(q-1)^2+4r]}{(q-r)^2}\le \frac{9q^2-9q+3}{2q^2}$

$\frac{9q^2-9q+3}{2q^2(1-q)}\ge \frac{(q-1)^2+4r}{(q-r)^2}=f(r)$ (2)

Because $f'(r)>0\Rightarrow f(r)$ is increasing function.

$\Rightarrow f(r)\le f(\frac{p}{9})=\frac{(q-1)^2+\frac{4q}{9}}{(\frac{8q}{9})^2}$

On the other hand:

$f(\frac{p}{9})=\frac{(q-1)^2+\frac{4q}{9}}{(\frac{8q}{9})^2}\le\frac{9q^2-9q+3}{2q^2(1-q)}$

$\Leftrightarrow q^3+q^2-q+\frac{5}{27}\ge 0$

$\Leftrightarrow (q-\frac{1}{3})^2(q+\frac{5}{3})\ge 0$ (true)

$\Rightarrow (2). True\Rightarrow (1). True!$

$\Rightarrow Q.E.D$

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#3 h Sep 19, 2010, 5:19 AM • 2 Y

Y by Adventure10, Mango247

minhhoang wrote:

Let a,b,c be positive real numbers. Prove that:

$\frac{1}{{{a^2} + ab + {b^2}}} + \frac{1}{{{b^2} + bc + {c^2}}} + \frac{1}{{{c^2} + ca + {a^2}}} \le \frac{{2({k^2} + k + 1)}}{{{a^2} + {b^2} + {c^2} + ab + bc + ca}}$

With $k = \frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ca}}$

similar for this:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022172#p2022172

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kuing

#4 h Sep 19, 2010, 5:31 AM • 1 Y

Y by Adventure10

kuing wrote:

minhhoang wrote:

Let a,b,c be positive real numbers. Prove that:

$\frac{1}{{{a^2} + ab + {b^2}}} + \frac{1}{{{b^2} + bc + {c^2}}} + \frac{1}{{{c^2} + ca + {a^2}}} \le \frac{{2({k^2} + k + 1)}}{{{a^2} + {b^2} + {c^2} + ab + bc + ca}}$

With $k = \frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ca}}$

similar for this:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022172#p2022172

hope this prove no miscalculation

$\frac{1}{{a^2 + ab + b^2 }} + \frac{1}{{b^2 + bc + c^2 }} + \frac{1}{{c^2 + ca + a^2 }} \le \frac{{2(k^2 + k + 1)}}{{a^2 + b^2 + c^2 + ab + bc + ca}}$

$\Leftarrow \frac{1}{{\left( {b + c} \right)^2 }} + \frac{1}{{\left( {c + a} \right)^2 }} + \frac{1}{{\left( {a + b} \right)^2 }} \le \frac{{3\left( {k^2 + k + 1} \right)}}{{2\left( {a^2 + b^2 + c^2 + ab + bc + ca} \right)}}$

$\Leftrightarrow \sum {\frac{{4ab + 4bc + 4ca}}{{\left( {b + c} \right)^2 }}} \le \frac{{6\left( {k^2 + k + 1} \right)\left( {ab + bc + ca} \right)}}{{a^2 + b^2 + c^2 + ab + bc + ca}}$

$\Leftrightarrow \sum {\frac{{4ab + 4ca - \left( {b - c} \right)^2 }}{{\left( {b + c} \right)^2 }}} \le \frac{{6\left( {k^2 + k + 1} \right)\left( {ab + bc + ca} \right)}}{{a^2 + b^2 + c^2 + ab + bc + ca}} - 3$

$\Leftarrow \sum {\frac{{4a}}{{b + c}}} \le \frac{{6\left( {k^2 + k + 1} \right)\left( {ab + bc + ca} \right)}}{{a^2 + b^2 + c^2 + ab + bc + ca}} - 3$

$\Leftrightarrow \sum {\frac{{2\left( {a - b} \right)^2 }}{{\left( {b + c} \right)\left( {c + a} \right)}}} \le \frac{{6\left( {k^2 + k + 1} \right)\left( {ab + bc + ca} \right)}}{{a^2 + b^2 + c^2 + ab + bc + ca}} - 9$

$\Leftrightarrow \sum {\frac{{2\left( {a - b} \right)^2 }}{{\left( {b + c} \right)\left( {c + a} \right)}}} \le \frac{{\left( {6\left( {a^2 + b^2 + c^2 } \right) + 3\left( {ab + bc + ca} \right)} \right)\sum {\left( {a - b} \right)^2 } }}{{2\left( {a^2 + b^2 + c^2 + ab + bc + ca} \right)\left( {ab + bc + ca} \right)}}$

$\Leftarrow \sum {\frac{{2\left( {a - b} \right)^2 }}{{\left( {b + c} \right)\left( {c + a} \right)}}} \le \frac{{9\sum {\left( {a - b} \right)^2 } }}{{4\left( {ab + bc + ca} \right)}}$

$\Leftrightarrow \sum {\frac{{\left( {a - b} \right)^2 }}{{9\left( {ab + bc + ca} \right) + 9c^2 }}} \le \frac{{\sum {\left( {a - b} \right)^2 } }}{{8\left( {ab + bc + ca} \right)}}$

obvious true.

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#5 h Dec 13, 2016, 4:23 PM • 1 Y

Y by Adventure10

minhhoang wrote:

Let a,b,c be positive real numbers. Prove that:

$\frac{1}{{{a^2} + ab + {b^2}}} + \frac{1}{{{b^2} + bc + {c^2}}} + \frac{1}{{{c^2} + ca + {a^2}}} \le \frac{{2({k^2} + k + 1)}}{{{a^2} + {b^2} + {c^2} + ab + bc + ca}}$

With $k = \frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ca}}$

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