a nice x,y,z ineq, in fact, holds for reals

by kuing, Dec 7, 2010, 4:31 AM

Let $x,y,z>0$ , prove that:

$\frac{(y+z-x)^2}{y^2+z^2}+\frac{(z+x-y)^2}{z^2+x^2}+\frac{(x+y-z)^2}{x^2+y^2} \geq \frac{3}{2}.$ (by xzlbq)

Prove(kuing):

$\begin{aligned} &\sum {\frac{{\left( {y + z - x} \right)^2 }} {{y^2 + z^2 }}} \geqslant \frac{3} {2}\\ \iff & \sum {\frac{{\left( {y + z} \right)^2 - 2x\left( {y + z} \right) + x^2 }} {{y^2 + z^2 }}} \geqslant \frac{3} {2} \\ \iff & \sum {\frac{{x^2 }} {{y^2 + z^2 }}} - \frac{3} {2} + \sum {\frac{{\left( {y + z} \right)\left( {y + z - 2x} \right)}} {{y^2 + z^2 }}} \geqslant 0 \\ \iff & \sum {\frac{{\left( {x + y} \right)^2 \left( {x - y} \right)^2 }} {{2\left( {y^2 + z^2 } \right)\left( {z^2 + x^2 } \right)}}} - \sum {\frac{{\left( {xy + yz + zx - z^2 } \right)\left( {x - y} \right)^2 }} {{\left( {y^2 + z^2 } \right)\left( {z^2 + x^2 } \right)}}} \geqslant 0 \\ \iff & \sum {\frac{{\left( {\left( {x + y} \right)^2 - 2\left( {xy + yz + zx - z^2 } \right)} \right)\left( {x - y} \right)^2 }} {{2\left( {y^2 + z^2 } \right)\left( {z^2 + x^2 } \right)}}} \geqslant 0 \\ \iff & \sum {\frac{{\left( {\left( {y - z} \right)^2 + \left( {z - x} \right)^2 } \right)\left( {x - y} \right)^2 }} {{2\left( {y^2 + z^2 } \right)\left( {z^2 + x^2 } \right)}}} \geqslant 0, \end{aligned}$
the last inequality is obvious true and we have done.
In fact, from this prove we can see that this inequaliy holds for any $x,y,z\in\mathbb{R}$ and $(x^2+y^2)(y^2+z^2)(z^2+x^2) \ne 0$ .
see also: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2107085#p2107085

This post has been edited 2 times. Last edited by kuing, Dec 7, 2010, 4:32 AM

six var ineq

by kuing, Dec 6, 2010, 4:31 AM

Let $a>x\ge 0,b>y\ge 0,c>z\ge 0$ , Prove

$\left((a+b+c)^3-(x+y+z)^3\right)\left(\frac{1}{a^3-x^3}+\frac{1}{b^3-y^3}+\frac{1}{c^3-z^3}\right)\ge 81.$
Prove(kuing):
Using Minkowski's Inequality, we have

$\begin{aligned} &\left( {\sqrt[3]{{x^3 }} + \sqrt[3]{{y^3 }} + \sqrt[3]{{z^3 }}} \right)^3 + \left( {\sqrt[3]{{a^3 - x^3 }} + \sqrt[3]{{b^3 - y^3 }} + \sqrt[3]{{c^3 - z^3 }}} \right)^3\\ \leqslant & \left( {\sqrt[3]{{a^3 }} + \sqrt[3]{{b^3 }} + \sqrt[3]{{c^3 }}} \right)^3 ,\end{aligned}$
and then we get that

$(a + b + c)^3 - (x + y + z)^3 \geqslant \left( {\sqrt[3]{{a^3 - x^3 }} + \sqrt[3]{{b^3 - y^3 }} + \sqrt[3]{{c^3 - z^3 }}} \right)^3 .$
So, Using Holder's Inequaltiy, we have

$\begin{aligned} & \left( {(a + b + c)^3 - (x + y + z)^3 } \right)\left( {\frac{1}{{a^3 - x^3 }} + \frac{1}{{b^3 - y^3 }} + \frac{1}{{c^3 - z^3 }}} \right) \\ \geqslant\; & \left( {\sqrt[3]{{a^3 - x^3 }} + \sqrt[3]{{b^3 - y^3 }} + \sqrt[3]{{c^3 - z^3 }}} \right)^3 \left( {\frac{1}{{a^3 - x^3 }} + \frac{1}{{b^3 - y^3 }} + \frac{1}{{c^3 - z^3 }}} \right) \\ \geqslant\; & \left( {1 + 1 + 1} \right)^4 \\ =\ & 81. \end{aligned}$
Done.

see also: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=380335
Remark, Minkowski's Inequality:
$a_i ,b_i > 0,p > \left( < \right)1 \implies \left( {\sum {a_i^p } } \right)^{\frac{1} {p}} + \left( {\sum {b_i^p } } \right)^{\frac{1} {p}} \geqslant \left( \leqslant \right)\left( {\sum {\left( {a_i + b_i } \right)^p } } \right)^{\frac{1} {p}} .$

This post has been edited 1 time. Last edited by kuing, Dec 6, 2010, 4:36 AM

remember another sqrt and cyc one

by kuing, Dec 5, 2010, 4:37 AM

Let $x,y,z>0$ , prove that

$\frac{x}{y}\cdot\frac{\sqrt{yz}}{y+z}+\frac{y}{z}\cdot\frac{\sqrt{zx}}{z+x}+\frac{z}{x}\cdot\frac{\sqrt{xy}}{x+y}\geq\frac{3}{2}.$ (by xzlbq)

Prove(kuing):
Let $x=a^2,y=b^2,z=c^2,a,b,c>0$ , then

$\begin{aligned} &\sum{\frac{x}{y}\cdot\frac{\sqrt{yz}}{y+z}} \geq \frac{3}{2}\\ \iff &\sum {\frac{{a^2 }}{{b^2 }}\cdot\frac{{bc}}{{b^2 + c^2 }}} \ge \frac{3}{2}\\ \iff &\sum {\frac{{\left( {\frac{{\sqrt {a^3 c} }}{{\sqrt b }}} \right)^2 }}{{a\left( {b^2 + c^2 } \right)}}} \ge \frac{3}{2}, \end{aligned}$
by Cauchy-Schwarz Inequality, we have

$\sum {\frac{{\left( {\frac{{\sqrt {a^3 c} }}{{\sqrt b }}} \right)^2 }}{{a\left( {b^2 + c^2 } \right)}}} \ge \frac{{\left( {\sum {\frac{{\sqrt {a^3 c} }}{{\sqrt b }}} } \right)^2 }}{{\sum {a\left( {b^2 + c^2 } \right)} }} = \frac{{\left( {\sum {a^2 c} } \right)^2 }}{{abc\sum {a\left( {b^2 + c^2 } \right)} }},$
thus, we just need to prove that

$2\left( {\sum {a^2 c} } \right)^2 \ge 3abc\sum {a\left( {b^2 + c^2 } \right)} .$ (nice one, mark it)
After expanding, easy to get that it is equivalent to

$\sum {c^2 \left( {2a^2 + ab} \right)\left( {a - b} \right)^2 } \ge 0,$
and we have done.

see also: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=377383
Kimpul have a nice prove in the link.

————2010-12-06 Edit————
another prove for the "mark it" one:
$\left( {\sum_{cyc} {ab^2 } } \right)^2 - 3abc\sum_{cyc} {a^2 b} = \frac{1}{2}\sum_{cyc} {(ab^2 - ca^2 )^2 } \geqslant 0$ $\implies \left( {\sum_{cyc} {ab^2 } } \right)^2 \geqslant 3abc\sum_{cyc} {a^2 b} .$ It's a stronger one because from this we can get that
$\begin{aligned} 2\left( {\sum_{cyc} {ab^2 } } \right)^2 & \geqslant 3abc\sum_{cyc} {a^2 b} + \left( {\sum_{cyc} {ab^2 } } \right)^2 \\ & \geqslant 3abc\sum_{cyc} {a^2 b} + 3abc\sum_{cyc} {ab^2 } \\ & = 3abc\sum_{cyc} {a\left( {b^2 + c^2 } \right)}. \end{aligned}$

This post has been edited 4 times. Last edited by kuing, Dec 6, 2010, 7:38 AM

sqrt and cyc one, how to prove the last ineq more classcal?

by kuing, Nov 24, 2010, 2:53 PM

Let $a,b,c > 0,a + b + c = 1$ , prove that :

$\frac{{\sqrt a }}{b} + \frac{{\sqrt b }}{c} + \frac{{\sqrt c }}{a} \ge 3\sqrt {\frac{a}{b} + \frac{b}{c} + \frac{c}{a}} .$
(by xzlbq, see also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=379036 , "h" one)

Prove(kuing):
Write the inequality in the homogeneous form, it's equivalent to

$\left( {a + b + c} \right)\left( {\frac{{\sqrt a }}{b} + \frac{{\sqrt b }}{c} + \frac{{\sqrt c }}{a}} \right)^2 \ge 9\left( {\frac{a}{b} + \frac{b}{c} + \frac{c}{a}} \right).$
Without loss of generality, assume $abc = 1$ , and let $a = \frac{{x^2 }}{{y^2 }},b = \frac{{y^2 }}{{z^2 }},c = \frac{{z^2 }}{{x^2 }},x,y,z > 0$ , then it's equivalent to

$\left( {\frac{{x^2 }}{{y^2 }} + \frac{{y^2 }}{{z^2 }} + \frac{{z^2 }}{{x^2 }}} \right)\left( {\frac{{z^2 x}}{{y^3 }} + \frac{{x^2 y}}{{z^3 }} + \frac{{y^2 z}}{{x^3 }}} \right)^2 \ge 9\left( {\frac{{z^2 x^2 }}{{y^4 }} + \frac{{x^2 y^2 }}{{z^4 }} + \frac{{y^2 z^2 }}{{x^4 }}} \right),$ or $\left( {x^2 y^4 + y^2 z^4 + z^2 x^4 } \right)\left( {x^5 y^4 + y^5 z^4 + z^5 x^4 } \right)^2 \ge 9x^4 y^4 z^4 \left( {x^6 y^6 + y^6 z^6 + z^6 x^6 } \right).$
Using Holder's Inequality, we have

$\left( {x^2 y^4 + y^2 z^4 + z^2 x^4 } \right)\left( {x^5 y^4 + y^5 z^4 + z^5 x^4 } \right)^2 \ge \left( {x^4 y^4 + y^4 z^4 + z^4 x^4 } \right)^3 ,$
so, let $u = x^2 y^2 ,v = y^2 z^2 ,w = z^2 x^2$ , then we just need to prove that

$\left( {u^2 + v^2 + w^2 } \right)^3 \ge 9uvw\left( {u^3 + v^3 + w^3 } \right). \qquad (*)$
Let $u + v + w = p,uv + vw + wu = q,uvw = r$ , then (*) is equivalent to

$f\left( r \right) = - 27r^2 - 9p\left( {p^2 - 3q} \right)r + \left( {p^2 - 2q} \right)^3 \ge 0,$
obviously, $f(r)$ is decreasing of $r$ , so we just need to check that (*) holds when two of $u,v,w$ are equal. Let $u = v = tw$ , then $t>0$ and (*) is equivalent to

$\left( {8t^4 - 2t^3 + 2t + 1} \right)\left( {t - 1} \right)^2 \ge 0,$ or $\left( {2t\left( {t + 1} \right)\left( {t - 1} \right)^2 + 6t^4 + 2t^2 + 1} \right)\left( {t - 1} \right)^2 \ge 0,$
obvious true, done.

This post has been edited 3 times. Last edited by kuing, Nov 24, 2010, 3:01 PM

sqrt and frac, own

by kuing, Nov 15, 2010, 2:30 PM

(kuing) Let $x,y,z>0$ , then the following inequality holds:

$\frac{{(\sqrt x + \sqrt y + \sqrt z )^2 }}{{\sqrt {x(y + z)} + \sqrt {y(z + x)} + \sqrt {z(x + y)} }} \ge 2\sqrt {\frac{{(x + y + z)(xy + yz + zx)}}{{(x + y)(y + z)(z + x)}}} .$

My prove:
Let $x = a^2 ,y = b^2 ,z = c^2 ,a,b,c > 0$ , then the inequality can be written as

$\frac{{\left( {\sum a } \right)^4 }}{{\left( {\sum {a\sqrt {b^2 + c^2 } } } \right)^2 }} \ge \frac{{4\sum {a^2 } \sum {a^2 b^2 } }}{{\prod {\left( {a^2 + b^2 } \right)} }},$ or
$\frac{{\left( {\sum a } \right)^2 }}{{\left( {\sum {\frac{a}{{a + b + c}}\sqrt {b^2 + c^2 } } } \right)^2 }} \ge \frac{{4\sum {a^2 } \sum {a^2 b^2 } }}{{\prod {\left( {a^2 + b^2 } \right)} }}.$
By Jensen's Inequality, we have

$\sum {\frac{a}{{a + b + c}}\sqrt {b^2 + c^2 } } \le \sqrt {\sum {\frac{{a\left( {b^2 + c^2 } \right)}}{{a + b + c}}} } = \sqrt {\frac{{\sum {a\left( {b^2 + c^2 } \right)} }}{{a + b + c}}} ,$
(in fact, this step can be written as Cauchy-Schwarz)
so we just need to prove that

\[
\begin{align*}
 &\frac{{\left( {\sum a } \right)^3 }}{{\sum {a\left( {b^2  + c^2 } \right)} }} \ge \frac{{4\sum {a^2 } \sum {a^2 b^2 } }}{{\prod {\left( {a^2  + b^2 } \right)} }} \\ 
  \iff & \frac{{\left( {\sum a } \right)^3 }}{{4\sum {a\left( {b^2  + c^2 } \right)} }} \ge \frac{{\prod {\left( {a^2  + b^2 } \right)}  + a^2 b^2 c^2 }}{{\prod {\left( {a^2  + b^2 } \right)} }} \\ 
  \iff & \frac{{\left( {\sum a } \right)^3 }}{{4\sum {a\left( {b^2  + c^2 } \right)} }} - 1 \ge \frac{{a^2 b^2 c^2 }}{{\prod {\left( {a^2  + b^2 } \right)} }} \\ 
 \iff & \frac{{\sum {a^3 }  - \sum {a\left( {b^2  + c^2 } \right)}  + 6abc}}{{4\sum {a\left( {b^2  + c^2 } \right)} }} \ge \frac{{a^2 b^2 c^2 }}{{\prod {\left( {a^2  + b^2 } \right)} }}. 
 \end{align*}
\]

By Schur's Inequaliy, we have

$\sum {a^3} - \sum {a\left( {b^2 + c^2 } \right)} + 6abc \ge 3abc,$
so we just need to prove that

\[
\begin{align*}
& \frac{{3abc}}{{4\sum {a\left( {b^2  + c^2 } \right)} }} \ge \frac{{a^2 b^2 c^2 }}{{\prod {\left( {a^2  + b^2 } \right)} }} \\ 
  \iff & 3\prod {\left( {a^2  + b^2 } \right)}  \ge 4abc\sum {a\left( {b^2  + c^2 } \right)}  \\ 
  \iff & \sum {c^2 (3a^2  + 2ab + 3b^2 )(a - b)^2 }  \ge 0,
 \end{align*}
\]

obvious true. The prove is completed.

see also: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=375071, in the topic, you can see that can_hang2007's prove is simpler and nicer than me.

Click to reveal hidden text

This post has been edited 3 times. Last edited by kuing, Nov 15, 2010, 2:39 PM

k(ab+bc+ca)>a^2+b^2+c^2 with triangle

by kuing, Nov 11, 2010, 8:16 AM

(1), If $a,b,c$ be the sides of a triangle, then we have

$2\left( {ab + bc + ca} \right) > a^2 + b^2 + c^2 ,$
and the coefficient $2$ is $\min$ .

(2), If $a,b,c>0$ and

$\frac{6}{5}\left( {ab + bc + ca} \right) > a^2 + b^2 + c^2 ,$
then $a,b,c$ can be the sides of a triangle, and the coefficient $\frac{6}{5}$ is $\max$ .

Click to reveal hidden text

This post has been edited 3 times. Last edited by kuing, Nov 11, 2010, 8:27 AM

another a^(1/2)+b^(1/2)+c^(1/2)>=ab+bc+ca

by kuing, Nov 9, 2010, 4:45 AM

(Proposed by Vasc) Let $a,b,c$ be nonnegative real numbers such that $(a+b)(b+c)(c+a)=8$ . Prove that
$\sqrt a+\sqrt b+\sqrt c \ge ab+bc+ca.$

My prove:
$\Leftrightarrow \left( {\sqrt a + \sqrt b + \sqrt c } \right)^2 \ge \frac{{8\left( {ab + bc + ca} \right)^2 }}{{\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)}} \\ \Leftrightarrow \frac{{\left( {\sqrt a + \sqrt b + \sqrt c } \right)^2 \left( {a + b + c} \right)}}{{8\left( {ab + bc + ca} \right)}} \ge \frac{{\left( {a + b + c} \right)\left( {ab + bc + ca} \right)}}{{\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)}} \\ \Leftrightarrow \frac{{\left( {\sqrt a + \sqrt b + \sqrt c } \right)^2 \left( {a + b + c} \right)}}{{8\left( {ab + bc + ca} \right)}} \ge \frac{{\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) + abc}}{{\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)}} \\ \Leftrightarrow \frac{{\left( {\sqrt a + \sqrt b + \sqrt c } \right)^2 \left( {a + b + c} \right)}}{{8\left( {ab + bc + ca} \right)}} - 1 \ge \frac{{abc}}{{\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)}}$
$(a = x^2 ,b = y^2 ,c = z^2 ,x,y,z > 0)$
$\Leftrightarrow \frac{{xyz(x + y + z) + \sum {\left( {x(x + 3y + 3z)(x - y)(x - z)} \right)} }}{{8\sum {x^2 y^2 } }} \ge \frac{{x^2 y^2 z^2 }}{{\prod {\left( {x^2 + y^2 } \right)} }} \\ \Leftarrow \frac{{x + y + z}}{{8\sum {x^2 y^2 } }} \ge \frac{{xyz}}{{\prod {\left( {x^2 + y^2 } \right)} }} \\ \Leftarrow \frac{1}{{8\sum {x^2 y^2 } }} \ge \frac{{x^2 + y^2 + z^2}}{{9\prod {\left( {x^2 + y^2 } \right)} }} \\ \Leftrightarrow 9(a+b)(b+c)(c+a)\ge 8(a+b+c)(ab+bc+ca)$
which is obvious true by Am-Gm

other proves can see also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=375405 they are all great.

This post has been edited 3 times. Last edited by kuing, Nov 9, 2010, 4:50 AM

a^(1/3)+b^(1/3)+c^(1/3)>=ab+bc+ca

by kuing, Nov 2, 2010, 7:42 AM

$a,b,c \ge 0$ such that $a+b+c=3$ , prove that

$\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} \ge ab + bc + ca.$
prove:
WOLG, assume $c = \min \left\{ {a,b,c} \right\}$ .

${\rm{Case}}\;1 : c < \frac{1}{6}$

easy to prove $\sqrt[3]{c} \ge 3c$ , then we get

$\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} \ge \sqrt[3]{a} + \sqrt[3]{b} + 3c = \sqrt[3]{a} + \sqrt[3]{b} + c^2 + bc + ca,$
so we just need to prove that

$\sqrt[3]{a} + \sqrt[3]{b} \ge ab.$
setting $t = \sqrt[6]{{ab}}$ , by Am-Gm Inequality, we have

$\sqrt[3]{a} + \sqrt[3]{b} - ab \ge 2t - t^6 = t\left( {2 - t^5 } \right)\;{\rm{and}}\;0 \le t^5 \le \left( {\frac{{a + b}}{2}} \right)^{\frac{5}{3}} \le \left( {\frac{3}{2}} \right)^{\frac{5}{3}} ,$
but $8 - \left( {\frac{3}{2}} \right)^5 = \frac{{13}}{{32}} > 0$ gives $t\left( {2 - t^5 } \right) \ge 0$ , i.e. $\sqrt[3]{a} + \sqrt[3]{b} \ge ab$ .
so in this case, inequality is true;

${\rm{Case}}\;2:c \ge \frac{1}{6}$

we have

\[
\begin{align*}
& \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} \ge ab + bc + ca \\ 
  \iff &\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} \ge \frac{{\left( {a + b + c} \right)^2  - \left( {a^2  + b^2  + c^2 } \right)}}{2} \\ 
  \iff &2\left( {\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c}} \right) + a^2  + b^2  + c^2  \ge 9 \\ 
  \iff &\sum\limits_{cyc} {\left( {2\sqrt[3]{a} + a^2  - \frac{{8a + 1}}{3}} \right)}  \ge 0 \\ 
  \iff &\sum\limits_{cyc} {\left( {\frac{1}{3}\left( {\sqrt[3]{a} - 1} \right)^2 \left( {3\sqrt[3]{{a^4 }} + 9\sqrt[3]{{a^2 }} + 4\sqrt[3]{a} + 6a - 1} \right)} \right)}  \ge 0,  
 \end{align*}
\]

obvious true because $a,b,c\ge\frac{1}{6}$ , so in this case, the inequality also true.
the prove is completed. done!
see also: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=367619
remark

This post has been edited 3 times. Last edited by kuing, Nov 2, 2010, 7:46 AM

ask a latex problem

by kuing, Oct 30, 2010, 5:01 AM

(1)
$\implies$ is \implies
$\Longrightarrow$ is \Longrightarrow
but
$A \implies B$ is A \implies B
$A \Longrightarrow B$ is A \Longrightarrow B
see the different?
similar:
(2)
$\iff$ is \iff
$\Longleftrightarrow$ is \Longleftrightarrow
but
$A \iff B$ is A \iff B
$A \Longleftrightarrow B$ is A \Longleftrightarrow B

we see that $\Longrightarrow$ and $\Longleftrightarrow$ have a "bigger spaces" style
and I think it should have the (3) one -- bigger spaces of $\Longleftarrow$ , but I don't know the command of this, how can tell me?

This post has been edited 3 times. Last edited by kuing, Oct 30, 2010, 5:43 AM

two of find k_max, \sqrt{9+6\sqrt{3}}

by kuing, Oct 23, 2010, 1:13 PM

(1), Find $k_{\max}$ such that for all $a,b,c \ge 0$ the following inequality holds:

$a^3 + b^3 + c^3 \ge k\left| {a - b} \right|\left| {b - c} \right|\left| {c - a} \right| + 3abc.$
(2), Find $k_{\max}$ such that for all $a,b,c \ge 0,(a-b)(b-c)(c-a)\ne0$ the following inequality holds:

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

Solution:

(1), Without loss of generality, we can assume that $a\ge b\ge c$ , if two of $a,b,c$ are equal, then the inequality is obvious true for any $k\in\mathbb{R}$ . Next, let's assume $a>b>c$ and $b=c+x,a=c+x+y,x,y>0$ , then

\[
\begin{align*}
& a^3  + b^3  + c^3  \ge k\left| {a - b} \right|\left| {b - c} \right|\left| {c - a} \right| + 3abc. \\ 
  \Leftrightarrow &\frac{1}{2}\left( {a + b + c} \right)\left( {\left( {a - b} \right)^2  + \left( {b - c} \right)^2  + \left( {c - a} \right)^2 } \right) \ge k\left| {a - b} \right|\left| {b - c} \right|\left| {c - a} \right| \\ 
  \Leftrightarrow &\frac{1}{2}\left( {3c + 2x + y} \right)\left( {x^2  + y^2  + \left( {x + y} \right)^2 } \right) \ge kxy\left( {x + y} \right) \\ 
  \Leftrightarrow & k \le \frac{{\left( {3c + 2x + y} \right)\left( {x^2  + y^2  + xy} \right)}}{{xy\left( {x + y} \right)}},
 \end{align*}
\]

so we just need to find the minimum of the function

$f\left( {x,y,c} \right) = \frac{{\left( {3c + 2x + y} \right)\left( {x^2 + y^2 + xy} \right)}}{{xy\left( {x + y} \right)}}.$
Let $y=tx,t>0$ , we have

$f\left( {x,y,c} \right) \ge f\left( {x,y,0} \right) = g\left( t \right) = \frac{{\left( {t^2 + t + 1} \right)\left( {t + 2} \right)}}{{t\left( {t + 1} \right)}},$
we need to find the minimum of the function $g\left( t \right)$ where $t>0$ ;

(2), We just need to find the minimum of the function

$f\left( {a,b,c} \right) = \left( {\frac{1}{{\left( {a - b} \right)^2 }} + \frac{1}{{\left( {b - c} \right)^2 }} + \frac{1}{{\left( {c - a} \right)^2 }}} \right)\left( {a + b + c} \right)^2.$
Without loss of generality, we can assume that $a>b>c$ and $b=c+x,a=c+x+tx,x,t>0$ , then

\[
\begin{align*}
 f\left( {a,b,c} \right)& = f\left( {c + x + tx,c + x,c} \right) \\ 
 & = \frac{{\left( {t^2  + t + 1} \right)^2 \left( {3c + tx + 2x} \right)^2 }}{{t^2 x^2 \left( {t + 1} \right)^2 }} \\ 
 & \ge \frac{{\left( {t^2  + t + 1} \right)^2 \left( {tx + 2x} \right)^2 }}{{t^2 x^2 \left( {t + 1} \right)^2 }} \\ 
 & = \frac{{\left( {t^2  + t + 1} \right)^2 \left( {t + 2} \right)^2 }}{{t^2 \left( {t + 1} \right)^2 }}\\
 & = \left(g(t)\right)^2
 \end{align*}
\]

so we also need to find the minimum of the (1)'s function $g(t)$ where $t>0$ .

But how to get the function minimum? see also: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2054717#p2054717
#7 gives the minimum is $\sqrt{9+6\sqrt{3}}$ , so the answer is
(1), $k_{\max}=\sqrt{9+6\sqrt{3}}$ ; (2), $k_{\max}=9+6\sqrt{3}$ .
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This post has been edited 8 times. Last edited by kuing, Oct 23, 2010, 1:32 PM

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Hello! $(A~\Rightarrow~B)~\Rightarrow~(\not B~\Rightarrow~\not A)$
by MaxSteinberg, Dec 4, 2014, 7:04 PM
???????????? hahah
by RobRoobiks, Nov 4, 2010, 3:17 PM
hey kuing, did you receive my message?
by dyta, Nov 3, 2010, 5:45 PM
oh, pxchg1200, here maybe must write in english, if not, will "????????????"...
by kuing, Oct 28, 2010, 9:02 AM
??????????????????
by pxchg1200, Oct 27, 2010, 1:24 AM
Oh!My Dear!do you often deal with so many complex identity translations?
by Lagrenge, Oct 23, 2010, 5:48 AM
sea rover, that's just Am-Gm ...
by kuing, Oct 23, 2010, 3:49 AM
??????????????????
$(b(c+a)+ca)((c+a)^2-ca) \le \frac{(c+a)^2(a+b+c)^2}{4}$
???????????????????????????
by sea rover, Oct 23, 2010, 3:29 AM
Thanks a lot, keep up your nice blog.
by Potla, Oct 22, 2010, 8:59 PM
WELL-DONE
by sea rover, Oct 22, 2010, 12:13 PM
thanks you very much!
by BaronShadeNight, Oct 12, 2010, 3:57 PM
welcome welcome very welcome!
by kuing, Sep 6, 2010, 1:23 PM

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