Chebyshev Inequality

Original Post.

A direct consequence of rearrangement, Chebyshev is a nice inequality that is able to deduce nicer results and can be extremely powerful sometimes too. This can be proved by rearrangement so this holds for all real numbers.
Statement
We already know it from our previous post; if a_i and b_i are similarly sorted sequences then
\sum_{i = 1}^n a_ib_i\geq \frac1n\sum a_i \sum b_i
And if these sequences are oppositely sorted then
\sum_{i = 1}^n a_ib_i\leq \frac1n\sum a_i \sum b_i
Since we already know it’s proof; I will directly go to applications of this inequality in various problems. We have already proven Iran 1996 using Chebyshev’s inequality. I have proved an inequality in Mathematical reflections this issue using simple Chebyshev; and I will post the inequality and my solution using Chebyshev after the deadline (1st November 2009) :) .We also have a little introduction to Some applications because of my previous post.So now I will have a short introduction to this inequality by deducing some known easy inequalities at first.
:)
Nessbitt’s inequality
We will prove for a,b,c > 0 we have \sum \frac a{b + c}\geq \frac32 in another way.
We have, the sequences a,b,c and \frac1{b + c}; \frac1{c + a}; \frac1{a + b} are similarly sorted so from Chebyshev;
\sum \frac a {b + c}\geq \frac 13 (a + b + c)\left(\frac1{b + c} + \frac1{c + a} + \frac1{a + b}\right)
\ge \frac 13(a + b + c) \left(\frac9{2(a + b + c)} = \frac 32 (From Cauchy Schwartz inequality)

2. (Problem by Darij Grinberg) For a,b,c > 0 prove that
\sum\frac {a}{(b + c)^{2}}\geq{{a + b + c}\over 4}

Since sequences a,b,c and \frac {1}{(b + c)^{2}},\frac {1}{(c + a)^{2}},\frac {1}{(a + b)^{2}} are similarly sorted hence from Chebyshev;
\sum\frac {a}{(b + c)^{2}}\geq\frac {1}3\sum a\sum\frac {1}{(b + c)^{2}}
\geq\frac {1}3\sum a{{9}\over{4(ab + bc + ca)}}(from Iran 1996)
\geq\frac {a + b + c}{4}\frac {3}{ab + bc + ca}\geq{{a + b + c}\over{4}}
Since ab + bc + ca\leq{{(a + b + c)^{2}}\over{3}}}\leq 3
So we are done.

Applications
\boxed{1}Let a,b,c denote the lengths of the three sides opposite to the three angles A,B,C of a triangle.
Show that:
\frac {\pi}{3}\leq\frac {aA + bB + cC}{a + b + c}\leq\frac {\pi}{2}
Solution
Let a \geq b \geq c so A \geq B \geq C. By Chebyshev inequality we have:

\frac {aA + bB + cC}{a + b + c}\geq \frac {(\frac {a + b + c}{3})(A + B + C)}{a + b + c} = \frac {A + B + C}{3} = \frac {\pi}...

For the right inequality, note that:
2aA + 2bB + 2cC\leq\pi(a + b + c)
2aA + 2bB + 2cC\leq aA + bA + cA + aB + bB + cB + aC + bC + cC
A(b + c - a) + B(c + a - b) + C(a + b - c)\geq0 which is true for a,b,c sides of a triangle.

\boxed{2}For a,b,c\ge0 Prove that \frac {a^{3}}{bc} + \frac {b^{3}}{ac} + \frac {c^{3}}{ab}\ge a + b + c

Solution

Sequences a,b,c and \frac1{bc}; \frac1{ac}; \frac1{ab} are similarly sorted.Hence from Chebyshev we have

LHS \geq \frac {1}{3}(a + b + c) \sum_{cyc}{\frac {a^2}{bc} \geq \frac {1}{3}(a + b + c)\frac {(a + b + c)^2}{ab + bc + ca} \... \Box
\boxed{3}Let a,b,c > 0 ; prove that :
\frac {(b + c)(b^2 + c^2)}{a} + \frac {(c + a)(c^2 + a^2)}{b} + \frac {(a + b)(a^2 + b^2)}{c}\geq 4(a^2 + b^2 + c^2)
Solution
Observe that sequences \frac {b + c}{a}; \frac {c + a}{b}; \frac {a + b}{c} and (b^2 + c^2); (c^2 + a^2); (a^2 + b^2) are similarly sorted (verify :) ). Hence we can apply Chebyshev on these to obtain:
\sum\frac {b + c}{a}\cdot (b^2 + c^2)\geq \frac 13 \sum \frac {b + c}{a}\sum (a^2 + b^2)

\geq \frac 13 \cdot 6 \cdot 2(a^2 + b^2 + c^2) = 4(a^2 + b^2 + c^2)}
Because from AM-GM we have \frac {b + c}{a} = \sum_{sym}\frac ab \geq 6.
Hence we are done :) \Box
\boxed{4} Let a;b;c be positive real numbers .Prove that:
\frac {a^{3} + b^{3} + c^{3}}{a + b + c} + \frac {a^{3} + b^{3} + d^{3}}{a + b + d} + \frac {a^{3} + c^{3} + d^{3}}{a + c + d...
Solution
By Chebyshev,we have
\frac {a^{3} + b^{3} + c^{3}}{a + b + c} \geq \frac {a^{2} + b^{2} + c^{2}}{3}.
Summing up these inequalities,we get the desired result.\Box
\boxed{5}Let a,b,c be positive reals such that these satisfy a + b + c = 3,
\frac {a}{a + 2bc} + \frac {b}{b + 2ac} + \frac {c}{c + 2ab} \leq 1
Solution
WLOG assume a\ge b \ge c
Because : (a^2 + b^2 + c^2) = \frac {1}{3}(a + b + c)(a^2 + b^2 + c^2) \ge 3abc
Hence:
\frac {a}{a + 2bc} + \frac {b}{b + 2ac} + \frac {c}{c + 2ab} - 1
= \frac {2(a - bc)}{a + 2bc} + \frac {2(b - ca)}{b + 2ac} + \frac {2(c - ab)}{c + 2ab}
= \frac {2(a^2 - abc)}{a^2 + 2abc} + \frac {2(b^2 - abc)}{b^2 + 2abc} + \frac {2(c^2 - abc)}{c^2 + 2abc}
\ge \frac {1}{3}.2(a^2 + b^2 + c^2 - 3abc)\left( \frac {1}{a^2 + 2abc} + \frac {1}{b^2 + 2abc} + \frac {1}{c^2 + 2abc}\right) (Using Chebyshev’s inequality)
\ge 0. QED \Box
\boxed{6}(Serbia 2008)Let a, b, c be positive real numbers such that a + b + c = 1. Prove that
\frac {1}{bc + a + \frac {1}{a}} + \frac {1}{ac + b + \frac {1}{b}} + \frac {1}{ab + c + \frac {1}{c}} \leq \frac {27}{31}.

Solution
We have the equivalent inequality :
\sum {\frac {9a^2 + 9abc + 9 - 31a}{a^2 + abc + 1}} \geq 0
Now we have the comment:
Because of a + b + c = 1 so that 0\leq a,b,c \leq 1
Without loss of generality, we assume a \geq b \geq c
But we also have the two monotonous sequences ( verify :) )
(a)9( a^2 + abc + 1) - 31a \leq 9(b^2 + abc + 1) - 31b \leq 9(c^2 + abc + 1) - 31c ( Note that 9( a + b) \leq 31)
(b) \frac {1}{ a^2 + abc + 1} \leq \frac {1}{ b^2 + abc + 1} \leq \frac {1}{ c^2 + abc + 1}
Applying Chebyshev inequality for two sequences above, we have
\sum{\frac {9(a^2 + abc + 1) - 31a}{ a^2 + abc + 1}}

\geq \sum{\frac {9(a^2 + abc + 1) - 31a}{ 3}}. \sum{\frac {1}{ a^2 + abc + 1}}
But it is easy to check by homogeneous method the following inequality:
a^2 + b^2 + c^2 + 3abc \geq \frac {4}{9} \Box
\boxed{7}Let a,b,c > 0, prove that: a^ab^bc^c\geq (abc)^{{{a + b + c}\over {3}}}
Solution
a^ab^bc^c\ge (abc)^{\frac {a + b + c}{3}}\Longleftrightarrow a^{3a}b^{3b}c^{3c}\ge (abc)^{a + b + c}

\Longleftrightarrow 3a\ln a + 3b\ln b + 3c\ln c\ge (a + b + c)(\ln a + \ln b + \ln c), which follows from Chebyshev’s inequality .
\Box
\boxed{8}. For a,b,c > 0, show that
\sum_{\text{cyc}} \frac {a^3}{a^2 + ab + b^2} \ge\frac {a + b + c}{3}

Solution
Because \sum {\frac {{{a^3}}}{{{a^2} + ab + {b^2}}}} = \sum {\frac {{{b^3}}}{{{a^2} + ab + {b^2}}}}
so,LHS = \frac 12 \sum {\frac {{{a^3} + {b^3}}}{{{a^2} + ab + {b^2}}}} \ge \frac {2}{3}\sum {\frac {{{a^3} + {b^3}}}{{{a^2} + {b...(by Chebyshev’s inequality, why? )
\boxed{9}Find the maximum value of following expression
A = \frac {x}{x^2 + yz} + \frac {y}{y^2 + zx} + \frac {z}{z^2 + xy} İf x,y,z are positive real numbers and x^2 + y^2 + z^2 = xyz
Solution(Honey_S)
Rewrite the inequality as:
\frac {x}{{x^2 + yz}} + \frac {{y^2 }}{{y^2 + zx}} + \frac {{z^2 }}{{z^2 + xy}} \le \frac {{x^2 + y^2 + z^2 }}{{2xyz}}

\Leftrightarrow \sum\limits_{cyc} {\left( {\frac {x}{{2yz}} - \frac {x}{{x^2 + yz}}} \right)} \ge 0

\Leftrightarrow \sum\limits_{cyc} {\frac {{x\left( {x^2 - yz} \right)}}{{yz\left( {x^2 + yz} \right)}}} \ge 0
Assume x \ge y \ge z then we have
x^2 - yz \ge y^2 - zx \Leftrightarrow \left( {x - y} \right)\left( {x + y + z} \right) \ge 0

\frac {{x^2 }}{{x^2 + yz}} \ge \frac {{y^2 }}{{y^2 + zx}} \Leftrightarrow \frac {{z\left( {x^3 - y^3 } \right)}}{{\left( {x^2...
Thus we have by Chebyshev inequality
\sum\limits_{cyc} {\frac {{x^2 \left( {x^2 - yz} \right)}}{{x^2 + yz}}} \ge \frac {1}{3}\left( {\sum\limits_{cyc} {\frac {{x^...
We finish our proof here :) \Box
\boxed{10}Prove that n^{2}(a_{1}^{3} + a_{2}^{3} + a_{3}^{3} + ... + a_{n}^{3})\geq(a_{1} + a_{2} + a_{3} + ... + a_{n})^{3} where n\in\mathbb N
Solution
Using Chebyshev’s inequality
\frac {(a_1 + a_2 + ... + a_n)}{n}\cdot \frac {(a_1 + a_2 + ... + a_n)}{n}\cdot \frac {(a_1 + a_2 + ... + a_n)}{n}

\leq \frac {(a_1 + a_2 + ... + a_n)}{n}\cdot \frac {a_1^2 + a_2^2 + ... + a_n^2}{n}

\leq \frac {a_1^3 + a_2^3 + ... + a_n^3}{n}
. \Box
\boxed{11} Prove that for a,b,c > 0 we always have:
\frac {2a}{(b + c)^2} + \frac {2b}{(c + a)^2} + \frac {2c}{(a + b)^2}\ge\frac {1}{a + b} + \frac {1}{b + c} + \frac {1}{c + a...
Solution
Assume WLOG that a \geq b \geq c. Then we have \frac {a}{b + c} \geq \frac {b}{a + c} \geq \frac {c}{a + b}, and \frac {1}{b + c} \geq \frac {1}{a + c} \geq \frac {1}{a + b}
Now by Chebyshev we have:
LHS \geq \frac {2}{3}\left(\frac {a}{b + c} + \frac {b}{a + c} + \frac {c}{a + b}\right)\left(\frac {1}{b + c} + \frac {1}{a ...
\geq \frac {2}{3}\cdot\frac {3}{2}\left(\frac {1}{b + c} + \frac {1}{a + c} + \frac {1}{a + b}\right)
= \frac {1}{b + c} + \frac {1}{a + c} + \frac {1}{a + b} \Box
\boxed{12}. Prove that for positive reals a,b,c, we have:
\frac {4a^2}{(b + c)^3} + \frac {4b^2}{(c + a)^3} + \frac {4c^2}{(a + b)^3}\ge\frac {a}{b^2 + c^2} + \frac {b}{c^2 + a^2} + \...

Solution
For positive reals x,y we have x^2 + y^2 \geq 2xy \Leftrightarrow 2(x^2 + y^2) \geq (x + y)^2 \Leftrightarrow (x + y)^3 \leq (x + y)2(x^2 + y^2), so we have:

\sum_{cyc}{\frac {4a^2}{(b + c)^3}} \geq \sum_{cyc}{\frac {4a^2}{2(b + c)(b^2 + c^2)}} = 2\sum_{cyc}{\frac {a}{b + c}\frac {a...

Assume
a \geq b \geq c so we have \frac {a}{b + c} \geq \frac {b}{a + c} \geq \frac {c}{b + a} and \frac {a}{b^2 + c^2} \geq \frac {b}{a^2 + c^2} \geq \frac {c}{b^2 + a^2}

By Chebyshev inequality we have now
2\sum_{cyclic}{\frac {a}{b + c}\frac {a}{b^2 + c^2}} \geq \frac {2}{3}\left(\frac {a}{b + c} + \frac {b}{a + c} + \frac {c}{b...
\geq \frac {2}{3}\cdot\frac {3}{2}\left(\frac {a}{b^2 + c^2} + \frac {b}{a^2 + c^2} + \frac {c}{b^2 + a^2}\right) = \left(\fr... . QED \Box
\boxed{13} Let a ;b and c are positive reals numbers. Show that:
\frac {a}{b + c + d} + \frac {b}{c + d + a} + \frac {c}{d + a + b} + \frac {d}{a + b + c} \geq \frac {4}{3}
Solution
(a,b,c,d) and \left(\frac {1}{b + c + d},\frac {1}{a + c + d},\frac {1}{a + b + d},\frac {1}{a + b + c}\right) are same ordered.
Hence, \frac {a}{b + c + d} + \frac {b}{c + d + a} + \frac {c}{d + a + b} + \frac {d}{a + b + c}
\geq\frac {1}{4}(a + b + c + d)\left(\frac {1}{b + c + d} + \frac {1}{a + c + d} + \frac {1}{a + b + d} + \frac {1}{a + b + c...
= \frac {1}{12}\sum_{cyc}(a + b + c)\sum_{cyc}\frac {1}{a + b + c}\geq\frac {1}{12}\cdot16 = \frac {4}{3} from Cauchy-Schwartz inequality. \Box
\boxed{14} Prove that for arbitrary non-negative real numbers a,b,c with sum 1 we have
\frac {x^{k + 2}}{x^{k + 1} + y^k + z^k} + \frac {y^{k + 2}}{y^{k + 1} + z^k + x^k} + \frac {z^{k + 2}}{z^{k + 1} + x^k + y^k...
Solution
Assume that x \ge y \ge z. It can be esily proved that
\frac {x^{k + 1}}{x^{k + 1} + y^k + z^k} \ge \frac {y^{k + 1}}{y^{k + 1} + z^k + x^k} \ge \frac {z^{k + 1}}{z^{k + 1} + x^k +...
z^{k + 1} + x^k + y^k \ge y^{k + 1} + z^k + x^k \ge x^{k + 1} + y^k + z^k.
Therefore applying Chebyshev’s Inequality we have
\quad\frac {x^{k + 2}}{x^{k + 1} + y^k + z^k} + \frac {y^{k + 2}}{y^{k + 1} + z^k + x^k} + \frac {z^{k + 2}}{z^{k + 1} + x^k ...
and
\quad\left( \frac {x^{k + 1}}{x^{k + 1} + y^k + z^k} + \frac {y^{k + 1}}{y^{k + 1} + z^k + x^k} + \frac {z^{k + 1}}{z^{k + 1}...

The rest is trivial. Hence we are done :) \Box
\boxed{15} Let a,b,c\in\mathbb R ^ +; Prove that :
\frac {1}{2}(\frac {b + c}{a} + \frac {c + a}{b} + \frac {b + a}{c}) \geq \frac {a + b + c}{\sqrt [3]{abc}}
Solution
From Chebyshev, since sequences b + c;c + a;c + a and \frac 1a; \frac 1b; \frac 1c are similarly sorted hence we have :
\frac {1}{2}\left(\frac {b + c}{a} + \frac {c + a}{b} + \frac {b + a}{c}\right)\ge \frac {1}{3}\left(\frac {1}{a} + \frac {1}...

\geq \frac 13 \cdot \frac {3}{\sqrt [3]{abc}}(a + b + c) = \frac {(a + b + c)}{\sqrt [3]{abc}}
(From AM-GM inequality).
Done. :) \Box
\boxed{16}Let x_1,x_2,...,x_n be distinct positive integers, then prove
a)x{_1}^2 + x{_2}^2 + ... + x{_n}^2 \ge \frac {2n + 1}{3}(x_1 + x_2 + ... + x_n)
Solution
Suppose that x_1 < x_2 < ... < x_n,hence x_i \geq i for all i
Put y_i = a_i - i,the inequality becomes:
\sum^{n}_{i = 1}b_i^2 + 2\sum^{n}_{i = 1}ib_i + \frac {n(n + 1)(2n + 1)}{6} \geq \frac {2n + 1}{3}\sum^{n}_{i = 1} b_i + \frac {n(n + 1)(2n + 1)}{6}
Note that b_1 \leq b_2 \leq ... \leq b_n
Then by Chebyshev inequality we have:
2\sum^{n}_{i = 1}ib_i \geq (n + 1)\sum ^{n}_{i = 1} b_i \geq \frac {2n + 1}{3}\sum^{n}_{i = 1} b_i
We are done. \Box
\boxed{17} Let a > b > c > 0 and abc = 1, Prove that:
\frac {9}{2}\leq\sum_{(a,b,c)}\frac {1}{a^{2}(b + c)}\cdot\sum_{(a,b,c)}ab\leq3\sum_{(a,b,c)}\frac {ab}{c^{2}(a + b)}
Solution
Let a = \frac {1}{x},b = \frac {1}{y},c = \frac {1}{z}. Then using xyz = 1, our inequality is equivalent to
\frac {9}{2}\le \sum\frac {x}{y + z}\cdot \sum x\le 3\sum \frac {x^{2}}{y + z}
For the left part, use \sum\frac {x}{y + z}\ge \frac {3}{2}\ge \frac {9}{2(x + y + z)}, which is true since x + y + z\ge 3.
Now we come to the right part.
Let x\le y\le z so that
x(x + z) \le y(y + z) \Leftrightarrow (x - y)(x + y + z) \ge 0,
Hence we can apply Chebyshev:
\sum\frac {x}{y + z}\cdot \sum x\le 3\sum \frac {x^{2}}{y + z}
QED. :) \Box
\boxed{18} Let be a,b and c non-negative reals numbers. Show that
\frac {b + c}{a^{2}} + \frac {a + c}{b^{2}} + \frac {a + b}{c^{2}}\geq\frac {2}{a} + \frac {2}{b} + \frac {2}{c}
Solution
rewrite the given inequality as:
(a + b + c)\left( \frac {1}{a^{2}} + \frac {1}{b^{2}} + \frac {1}{c^{2}}\right) \geq 3\left( \frac {1}{a} + \frac {1}{b} + \f...
after adding \frac {1}{a} to \frac {b + c}{a^{2}}, etc. Now the above inequality is evident by Chebyshev.
\Box
\boxed{19} Let x, y, z \ge 0 be real numbers. Prove that:
\frac {x^{3} + y^{3} + z^{3}}{3}\ge xyz + \frac {3}{4}|(x - y)(y - z)(z - x)| .

Solution
Rewrite the inequality as:
\frac {x^3 + y^3 + z^3 - 3xyz}{3}\geq \frac 94 |(x - y)(y - z)(z - x)|

\implies \left[(x - y)^2 + (y - z)^2 + (z - x)^2 \right]\left[(x + y) + (y + z) + (z + x)

\geq 9|(x - y)(y - z)(z - x)|
But we have; x + y\geq |x - y|; y + z\geq |y - z|; z + x\geq |z - x|. So we are only required to prove that :
\left[(x - y)^2 + (y - z)^2 + (z - x)^2\right]\left[|x - y| + |y - z| + |z - x|\right]

\geq 9|(x - y)(y - z)(z - x)|
Which is obvious from chebyshev’s inequality. :)
Equality holds iff x = y = z = 0 \Box
\boxed{20} If a,b,c > 0 , \sum \frac {a^{2}}{b + c}\geq \frac {\sqrt {3}}{2}\cdot \sqrt {a^{2} + b^{2} + c^{2}}\geq \frac {a + b + c}{2}
Solution
WLOG a\ge b\ge c. then from Chebyshev we have:
\sum\frac {a^{2}}{b + c}\ge \frac {3(a^{2} + b^{2} + c^{2})}{2(a + b + c)}
so we need to show that
\sqrt {3(a^{2} + b^{2} + c^{2})}\ge a + b + c
which is just the trivial inequality. Note that the right side also follows from the last inequality. \Box
\boxed{21}Let a,b,c be positive real numbers, show that
\frac1{a\sqrt {2(a^2 + bc)}} + \frac1{b\sqrt {2(b^2 + ca)}} + \frac1{c\sqrt {2(c^2 + ab)}}\ge\frac9{2(ab + bc + ca)}

Solution
Lemma
We claim that
\sum_{cyc}\frac {b + c}{(a + b)(a + c)} \ge \frac {3\sqrt {3}}{2\sqrt {ab + bc + ca}}
Hint for proof : You can use Iran 1996 Inequality. This remains as a challenge for the readers. \Delta
We have
\frac {1}{a\sqrt {2(a^{2} + bc)}} = \frac {\sqrt {b + c}}{\sqrt {2a}.\sqrt {(ab + ac)(a^2 + bc)}} \ge \frac {\sqrt {2(b + c)}...
Hence, it suffices to show that
\sum_{cyc}\frac {\sqrt {2}(b + c)}{\sqrt {a(b + c)}.(a + b)(a + c)} \ge \frac {9}{2(ab + bc + ca)}
Now, WLOG, we may assume that a \ge b \ge c and thus,
\left\{\begin{array} {cc} \frac {b + c}{(a + c)(a + b)} \le \frac {a + c}{(b + c)(b + a)} \le \frac {b + a}{(c + a)(c + b)} \...
Again, using Chebyshev inequality for these and our lemma above the proof is quite obvious.
_______________________
Now since the deadline for Mathematical Reflections is over (1st Nov. 2009) hence I will post the inequality and my solutions for it.
See my attachment, it is the same document that I sent to MR, but unfortunately got no reply……
You will also get a fine demonstration of using rearrangement nicely.
Please enjoy. (Also tell me if you like my solutions)
* Link for Mathematical reflections 2009 issue 5 submissions:
http://reflections.awesomemath.org/currentissue.html
_______________________
(To be updated)
References
1. A generalization of Chebyshev’s inequality. :http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244168

2. The “Four Solutions” Link.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s