it's equavilent $ 9 \sum ab + 1 \ leq 108abc$

Hong Quy wrote: it's equavilent $ 9 \sum ab + 1 \ leq 108abc$ Why :

Hong Quy I think you just want to spam Please do not particpate mathlinks People can think bad about Vietnamese

i don't spam. i use Cauchy $ \frac {a}{abc + a^2 + 1} = \frac {a}{abc + \frac {3a^2}{3} + \frac {27}{27}} \leq \frac {1}{31^2}(\frac {1}{bc} + \frac {9}{a} + 27^2)$ hence $ LHS \leq \frac {1}{31^2} ( \sum \frac {1}{bc} + 9\sum \frac {1}{a} + 3.27^2 ) \leq RHS$ it's equavilent with ineq $ 9 \sum ab + 1 \leq 108abc$

Hong Quy wrote: i don't spam. i use Cauchy $ \frac {a}{abc + a^2 + 1} = \frac {a}{abc + \frac {3a^2}{3} + \frac {27}{27}} \leq \frac {1}{31^2}(\frac {1}{bc} + \frac {9}{a} + 27^2)$ hence $ LHS \leq \frac {1}{31^2} ( \sum \frac {1}{bc} + 9\sum \frac {1}{a} + 3.27^2 ) \leq RHS$ it's equavilent with ineq $ 9 \sum ab + 1 \leq 108abc$ But $ 9\sum{ab}+1 \geq 108abc$

Can somebody post the solution? :

$ f"(x) = ( \frac{1}{bc+x+\frac{1}{x}})" = \frac{ 2(x^3+x^2-3x-1-bc)}{(xbc+x^2+1)^3} \leq 0$ $ \forall x \in [0;1]$ Use Jensen's ineq we have : $ f(a) +f(b) + f(c) \leq 3 f(\frac{a+b+c}{3}) =3f(\frac{1}{3}) = \frac{27}{31}$ seem it 's wrong ! but i don't sure

Hong Quy I think you just want to spam Please do not particpate mathlinks People can think bad about Vietnamese If you don 't sure please don't post It take many time to check a solution

Hong Quy wrote: $ f"(x) = ( \frac {1}{bc + x + \frac {1}{x}})" = \frac { 2(x^3 + x^2 - 3x - 1 - bc)}{(xbc + x^2 + 1)^3} \leq 0$ $ \forall x \in [0;1]$ Use Jensen's ineq we have : $ f(a) + f(b) + f(c) \leq 3 f(\frac {a + b + c}{3}) = 3f(\frac {1}{3}) = \frac {27}{31}$ seem it 's wrong ! but i don't sure Hong Quy, it's obviously wrong because $ x$ is function of $ b$ and $ c.$

alexilic wrote: Let $ a$, $ b$, $ c$ be positive real numbers such that $ a + b + c = 1$. Prove inequality: \[ \frac {1}{bc + a + \frac {1}{a}} + \frac {1}{ac + b + \frac {1}{b}} + \frac {1}{ab + c + \frac {1}{c}} \leqslant \frac {27}{31}. \] See my solution here

later, i have solution for this problem. use Schur's ineq with $ a+b+c =1$ we have : $ abc \geq 4(ab+bc+ca) -1$ (*) by other form , this ineq's equavilent with : $ \sum \frac{a}{abc +a^2+1} \leq \frac{27}{31}$ use (*) $ \to$ we need prove that : $ \sum (1-\frac{31a(a+b+c)}{\sum ab + 9a^2 +8} \leq 0$ or $ \sum \frac{(7a+8c+10b)(c-a) -(7a+8b+10c)(a-b)}{a^2 + k} \leq 0$ where $ k = \frac{4 \sum ab +8}{9}$ later, $ \sum (a-b)^2\frac{17ab+8(a62+b^2)+10c(a+b)+k }{(a^2+k)(b^2+k)} \geq 0$ occur equal when : $ a=b=c=1/3$

hello evry body , here is my solution ( hope it's right ) , it's very ugly the inequality is equivalent to : \[ \sum{\frac {1}{bc + a + \frac {1}{a}} - a} \leq \frac { - 4}{31} \] ( because $ a + b + c = 1$) \[ \sum{\frac {a^2(bc + a)}{abc + a^2 + 1}} \geq \frac {4}{31} \] \[ \sum{\frac { (a^2(bc + a)^2)}{(abc + a^2 + 1)(bc + a)}} \geq \frac {4}{31} \] but by cauchy shwarz : \[ LHS(\sum{(abc + a^2 + 1)(bc + a)}) \geq (\sum{a(bc + a)})^2 \] setting : $ r = abc,q = ab + ac + bc$ and since $ a + b + c = 1$ it's easy the check that: $ \sum{(abc + a^2 + 1)(bc + a)} = 2 - 2q + 5r + qr$ $ \sum{a(bc + a)} = 3r + 1 - 2q$ so we need to prove that: $ (3r + 1 - 2q)^2 - \frac {4(2 - 2q + 5r + qr)}{31} \geq 0$ or: $ 9r^2 + \frac {166r}{31} - \frac {376qr}{31} + \frac {23}{31} - \frac {116q}{31} + 4q^2 \geq 0$ now we put: $ f(r) = 9r^2 + \frac {166r}{31} - \frac {376qr}{31} + \frac {23}{31} - \frac {116q}{31} + 4q^2$ we have: $ f(r)' = 18r + \frac {166}{31} - \frac {376q}{31}$ it's easy to check that : $ f(r)'\geq0$ since $ q\leq \frac {1}{3}$ so f is an increasing function , and by shur : $ r\geq \frac {4q - 1}{9}$ thus :$ f(r)\geq f(\frac {4q - 1}{9}) = \frac {12q^2}{31} - \frac {28q}{31} + \frac {8}{31}$ hence it sufficies to prove , that : $ \frac {12q^2}{31} - \frac {28q}{31} + \frac {8}{31} \geq 0$ wich is equivalent to : $ \frac {4(3q - 1)(q - 2)}{31} \geq 0$ , wich is true since $ q\le\frac {1}{3}\leq 2$ @can_hang2007 : I really like your solution , is it a Intuition or a technique that I should learn? , thank you very much

Please , someone confirm my solution! thank you

anas wrote: hello evry body , here is my solution ( hope it's right ) , it's very ugly the inequality is equivalent to : \[ \sum{\frac {1}{bc + a + \frac {1}{a}} - a} \leq \frac { - 4}{31} \] ( because $ a + b + c = 1$) \[ \sum{\frac {a^2(bc + a)}{abc + a^2 + 1}} \geq \frac {4}{31} \] \[ \sum{\frac { (a^2(bc + a)^2)}{(abc + a^2 + 1)(bc + a)}} \geq \frac {4}{31} \] but by cauchy shwarz : \[ LHS(\sum{(abc + a^2 + 1)(bc + a)}) \geq (\sum{a(bc + a)})^2 \] setting : $ r = abc,q = ab + ac + bc$ and since $ a + b + c = 1$ it's easy the check that: $ \sum{(abc + a^2 + 1)(bc + a)} = 2 - 2q + 5r + qr$ $ \sum{a(bc + a)} = 3r + 1 - 2q$ so we need to prove that: $ (3r + 1 - 2q)^2 - \frac {4(2 - 2q + 5r + qr)}{31} \geq 0$ or: $ 9r^2 + \frac {166r}{31} - \frac {376qr}{31} + \frac {23}{31} - \frac {116q}{31} + 4q^2 \geq 0$ now we put: $ f(r) = 9r^2 + \frac {166r}{31} - \frac {376qr}{31} + \frac {23}{31} - \frac {116q}{31} + 4q^2$ we have: $ f(r)' = 18r + \frac {166}{31} - \frac {376q}{31}$ it's easy to check that : $ f(r)'\geq0$ since $ q\leq \frac {1}{3}$ so f is an increasing function , and by shur : $ r\geq \frac {4q - 1}{9}$ thus :$ f(r)\geq f(\frac {4q - 1}{9}) = \frac {12q^2}{31} - \frac {28q}{31} + \frac {8}{31}$ hence it sufficies to prove , that : $ \frac {12q^2}{31} - \frac {28q}{31} + \frac {8}{31} \geq 0$ wich is equivalent to : $ \frac {4(3q - 1)(q - 2)}{31} \geq 0$ , wich is true since $ q\le\frac {1}{3}\leq 2$ @can_hang2007 : I really like your solution , is it a Intuition or a technique that I should learn? , thank you very much ur solution's right and nice.

anas wrote: hello evry body , here is my solution ( hope it's right ) , it's very ugly the inequality is equivalent to : \[ \sum{\frac {1}{bc + a + \frac {1}{a}} - a} \leq \frac { - 4}{31} \] ( because $ a + b + c = 1$) \[ \sum{\frac {a^2(bc + a)}{abc + a^2 + 1}} \geq \frac {4}{31} \] \[ \sum{\frac { (a^2(bc + a)^2)}{(abc + a^2 + 1)(bc + a)}} \geq \frac {4}{31} \] but by cauchy shwarz : \[ LHS(\sum{(abc + a^2 + 1)(bc + a)}) \geq (\sum{a(bc + a)})^2 \] setting : $ r = abc,q = ab + ac + bc$ and since $ a + b + c = 1$ it's easy the check that: $ \sum{(abc + a^2 + 1)(bc + a)} = 2 - 2q + 5r + qr$ $ \sum{a(bc + a)} = 3r + 1 - 2q$ so we need to prove that: $ (3r + 1 - 2q)^2 - \frac {4(2 - 2q + 5r + qr)}{31} \geq 0$ or: $ 9r^2 + \frac {166r}{31} - \frac {376qr}{31} + \frac {23}{31} - \frac {116q}{31} + 4q^2 \geq 0$ now we put: $ f(r) = 9r^2 + \frac {166r}{31} - \frac {376qr}{31} + \frac {23}{31} - \frac {116q}{31} + 4q^2$ we have: $ f(r)' = 18r + \frac {166}{31} - \frac {376q}{31}$ it's easy to check that : $ f(r)'\geq0$ since $ q\leq \frac {1}{3}$ so f is an increasing function , and by shur : $ r\geq \frac {4q - 1}{9}$ thus :$ f(r)\geq f(\frac {4q - 1}{9}) = \frac {12q^2}{31} - \frac {28q}{31} + \frac {8}{31}$ hence it sufficies to prove , that : $ \frac {12q^2}{31} - \frac {28q}{31} + \frac {8}{31} \geq 0$ wich is equivalent to : $ \frac {4(3q - 1)(q - 2)}{31} \geq 0$ , wich is true since $ q\le\frac {1}{3}\leq 2$ @can_hang2007 : I really like your solution , is it a Intuition or a technique that I should learn? , thank you very much I have another solution, everybody can enjoy it! 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 ( you can check it) * $ 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)$ ** $ \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 homogenious method the following inequality: $ a^2 + b^2 + c^2 + 3abc \geq \frac {4}{9}$ An elementary proof! Can you agree with me?

sorry, i did not realize that

omerftekin wrote: This inequality is a special case of İlham's Cauchy. Putting n=31 in Famous (Meshuur) İlham's Cauchy we get the solution. Post please, the İlham's Cauchy inequality here. Thx!

cauchy schwarz bunyakovski inequalıty is specıal case of ılham's cauchy.are you serıous that you don't know this inequality.

omerftekin wrote: cauchy schwarz bunyakovski inequalıty is specıal case of ılham's cauchy.are you serıous that you don't know this inequality. Do you enjoy to post spam like this? All your posts are full of non-sense contents like that. Please stop to do that.

Let $ abc-ab-bc-ac+1=m$ and $ 2abc+2=n$. Inequality equilavents to $ \frac{-mn+m+n}{\frac{nm^2}{2}+1} \leq \frac{27}{31}$ $ \Longleftrightarrow$ $ 62(m+n) \leq n(27m^2+62m)+54$ $ n\leq \frac{56}{27}$ $ \Longrightarrow 62(m+n) \leq 62(m+\frac{56}{27})$. We will prove that $ 62(m+\frac{56}{27}) \leq n(27m^2+62m)+54$ By Schur's inequality $ (a+b+c)^3+9abc \geq 4(ab+bc+ac)(a+b+c)$ $ \Longrightarrow$ $ n \geq \frac{16-8m}{5}$ By $ AM$-$ GM$ $ (a+b)(b+c)(a+c) \leq \frac{8}{27}(a+b+c)^3= \frac{8}{27} \Longrightarrow$ $ m=abc-ab-bc-ac+1 \geq \frac{19}{27}$ $ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{(a+b+c)^2}>1 \Longrightarrow m<1 \Longrightarrow (2m+1)^2-\frac{292}{27}<0 \Longrightarrow$ $ (27m-19)(4m^2+4m-\frac{265}{27})\leq0 \Longrightarrow$ $ 108m^3+32m^2-341m+\frac{5035}{27}\leq0 \Longleftrightarrow$ $ 62(m+\frac{56}{27}) \leq \frac{16-8m}{5}.(27m^2+62m)+54 \leq n(27m^2+62m)+54$

alexilic wrote: Let $ a$, $ b$, $ c$ be positive real numbers such that $ a + b + c = 1$. Prove inequality: \[ \frac {1}{bc + a + \frac {1}{a}} + \frac {1}{ac + b + \frac {1}{b}} + \frac {1}{ab + c + \frac {1}{c}} \leqslant \frac {27}{31}. \] I think it's easy. Using Am-Gm, we are easily to prove the stronger: $ (\frac {1}{bc + a + \frac {1}{a}})^{\frac {10}{9}} + (\frac {1}{ca + b + \frac {1}{b}})^{\frac {10}{9}} + (\frac {1}{ab + c + \frac {1}{c}})^{\frac {10}{9}} \leq\ 3 (\frac {9}{31})^{\frac {10}{9}}$ (with the same conditon) With this stronger, all of the proofs above can't solve it.

How can you prove it? It seems like harder than original one. Could you please post that AM-GM solution?

nguoivn wrote: alexilic wrote: Let $ a$, $ b$, $ c$ be positive real numbers such that $ a + b + c = 1$. Prove inequality: \[ \frac {1}{bc + a + \frac {1}{a}} + \frac {1}{ac + b + \frac {1}{b}} + \frac {1}{ab + c + \frac {1}{c}} \leqslant \frac {27}{31}. \] I think it's easy. Using Am-Gm, we are easily to prove the stronger: $ (\frac {1}{bc + a + \frac {1}{a}})^{\frac {10}{9}} + (\frac {1}{ca + b + \frac {1}{b}})^{\frac {10}{9}} + (\frac {1}{ab + c + \frac {1}{c}})^{\frac {10}{9}} \leq\ 3 (\frac {9}{31})^{\frac {10}{9}}$ (with the same conditon) With this stronger, all of the proofs above can't solve it. I cannot find the solution either. Could you please elaborate a little?

Very sorry, my friend. I'll post my proof (but not now) because of some reason

nguoivn wrote: alexilic wrote: Let $ a$, $ b$, $ c$ be positive real numbers such that $ a + b + c = 1$. Prove inequality: \[ \frac {1}{bc + a + \frac {1}{a}} + \frac {1}{ac + b + \frac {1}{b}} + \frac {1}{ab + c + \frac {1}{c}} \leqslant \frac {27}{31}. \] I think it's easy. Using Am-Gm, we are easily to prove the stronger: $ (\frac {1}{bc + a + \frac {1}{a}})^{\frac {10}{9}} + (\frac {1}{ca + b + \frac {1}{b}})^{\frac {10}{9}} + (\frac {1}{ab + c + \frac {1}{c}})^{\frac {10}{9}} \leq\ 3 (\frac {9}{31})^{\frac {10}{9}}$ (with the same conditon) With this stronger, all of the proofs above can't solve it. Now, here is my proof:

Attachments:
nguoivn1.doc (46kb)

annn's solution is good.

anns's solution is good

alexilic wrote: Let $ a$, $ b$, $ c$ be positive real numbers such that $ a + b + c = 1$. Prove inequality: \[ \frac{1}{bc + a + \frac{1}{a}} + \frac{1}{ac + b + \frac{1}{b}} + \frac{1}{ab + c + \frac{1}{c}} \leqslant \frac{27}{31}.\] I only use the AG-GM and Cauchy-Schwarz: $LS=\sum\frac{a}{a(a+b)(a+c)+1}\\ =\sum\frac{a}{(1-b-c)(a+b)(a+c)+1}\\ =\sum\frac{a}{(a+b)(a+c)+1-(a+b)(b+c)(c+a)}$ By CauChy-Schwarz, conspicuous: $LS \le \sum\frac{a}{31^2}[\frac{12^2}{(a+b)(a+c)}+\frac{19^2}{1-(a+b)(b+c)(c+a)}]\\ =\sum\frac{12^2}{31^2}[\frac{a}{(a+b)(a+c)}]+\frac{19^2}{31^2}[\frac{a+b+c}{1-(a+b)(b+c)(c+a)}]=L$ On the other hand: $\sum\frac{a}{(a+b)(a+c)}=\frac{2(ab+bc+ca)}{(a+b)(b+c)(c+a)}\le \frac{9}{4}$ [because $(a+b)(b+c)(c+a)\ge \frac{8}{9}(ab+bc+ca)$] And: $\frac{1}{1-(a+b)(b+c)(c+a)}\le \frac{1}{1-[\frac{2(a+b+c)}{3}]^3}=\frac{27}{19}$ (AG-GM) So: $L\le \frac{12^2}{31^2}.\frac{9}{4}+\frac{19^2}{31^2}.\frac{27}{19}=\frac{27}{31}$ $\Rightarrow Q.E.D$

BaronShadeNight wrote: alexilic wrote: Let $ a$, $ b$, $ c$ be positive real numbers such that $ a + b + c = 1$. Prove inequality: \[ \frac{1}{bc + a + \frac{1}{a}} + \frac{1}{ac + b + \frac{1}{b}} + \frac{1}{ab + c + \frac{1}{c}} \leqslant \frac{27}{31}.\] I only use the AG-GM and Cauchy-Schwarz: $LS=\sum\frac{a}{a(a+b)(a+c)+1}\\ =\sum\frac{a}{(1-b-c)(a+b)(a+c)+1}\\ =\sum\frac{a}{(a+b)(a+c)+1-(a+b)(b+c)(c+a)}$ By CauChy-Schwarz, conspicuous: $LS \le \sum\frac{a}{31^2}[\frac{12^2}{(a+b)(a+c)}+\frac{19^2}{1-(a+b)(b+c)(c+a)}]\\ =\sum\frac{12^2}{31^2}[\frac{a}{(a+b)(a+c)}]+\frac{19^2}{31^2}[\frac{a+b+c}{1-(a+b)(b+c)(c+a)}]=L$ On the other hand: $\sum\frac{a}{(a+b)(a+c)}=\frac{2(ab+bc+ca)}{(a+b)(b+c)(c+a)}\le \frac{9}{4}$ [because $(a+b)(b+c)(c+a)\ge \frac{8}{9}(ab+bc+ca)$] And: $\frac{1}{1-(a+b)(b+c)(c+a)}\le \frac{1}{1-[\frac{2(a+b+c)}{3}]^3}=\frac{27}{19}$ (AG-GM) So: $L\le \frac{12^2}{31^2}.\frac{9}{4}+\frac{19^2}{31^2}.\frac{27}{19}=\frac{27}{31}$ $\Rightarrow Q.E.D$ very nice

Let $ a$, $ b$, $ c$ be positive real numbers such that $ a + b + c = 1$. Prove inequality: \[ \frac{1}{b^2 + a + \frac{1}{a}} + \frac{1}{c^2 + b + \frac{1}{b}} + \frac{1}{a^2+ c + \frac{1}{c}} \leqslant \frac{27}{31} \]\[ \frac{a}{b^2 + c + \frac{1}{c}} + \frac{b}{c^2 + a + \frac{1}{a}} + \frac{c}{a^2+b+ \frac{1}{b}} \leqslant \frac{9}{31} \]\[ \frac{a}{a^2+ b + \frac{1}{b}}+ \frac{b}{b^2 + c + \frac{1}{c}} + \frac{c}{c^2 + a+ \frac{1}{a}} \leqslant \frac{9}{31} \]

Rewrite $a=\frac{x}{3},b=\frac{y}{3},c=\frac{z}{3}$. \[ \frac{1}{bc + a + \frac{1}{a}} + \frac{1}{ac + b + \frac{1}{b}} + \frac{1}{ab + c + \frac{1}{c}} \leqslant \frac{27}{31}\iff \sum{\frac{x}{xyz+3x^2+27}}\overset{?}{\leq} \frac{3}{31}\]\[\iff \sum{1-\frac{31x}{xyz+3x^2+27}}=\sum{\frac{27+3x^2+xyz-31x}{xyz+3x^2+27}}\overset{?}{\geq} 0\]WLOG $x\geq y\geq z$ (the inequality is symmetric). We have $27+3x^2+xyz-31x\leq 27+3y^2+xyz-31y\leq 27+3z^2+xy-31z$ since $(31-3x-3y)(x-y)\geq 0$ and $(31-3y-3z)(y-z)\geq 0$. Also $xyz+3x^2+27\geq xyz+3y^2+27\geq xyz+3z^2+27$. Hence we can apply Chebyshev on decreasing sequences $(27+3x^2+xyz-31x,27+3y^2+xyz-31y,27+3z^2+xyz-31z)$ and $\frac{1}{xyz+3x^2+27},\frac{1}{xyz+3y^2+27},\frac{1}{xyz+3z^2+27})$. This gives \[\sum{\frac{27+3x^2+xyz-31x}{xyz+3x^2+27}}\geq \frac{1}{3}\sum{(27+3x^2+xyz-31x)}\sum{\frac{1}{xyz+3x^2+27}}\overset{?}{\geq} 0\]Let's show that $81+3\sum{x^2}+3xyz\overset{?}{\geq} 31\sum{x}=93$ or $\sum{x^2}+xyz\geq 4$. Claim: If $x+y+z=3$ for positive reals, then $x^2+y^2+z^2+xyz\geq 4$. Proof: First, we observe that if $max\{x,y,z\}\geq 2$, then $LHS\geq 4$. Suppose that $x,y,z<2$. Assume that $x,y$ are on different sides with respect to $1$. Let $x+y=p,xy=q$. Note that by our assumption, $p>1$. \[x^2+y^2+(3-x-y)^2+xy(3-x-y)\overset{?}{\geq} 4\iff p^2-2q+p^2-6p+9+3q-pq\overset{?}{\geq} 4\]\[\iff 2p^2-6p+q-pq+5\overset{?}{\geq} 0\iff q\overset{?}{\leq} \frac{2p^2-6p+5}{p-1}=2p-4+\frac{1}{p-1}\]\[2x+2y-4+\frac{1}{x+y-1}=(x+y-3)+(x+y-1)+\frac{1}{x+y-1}\geq (x+y-3)+2=x+y-1\overset{?}{\geq} xy\]The last part follows by $(x-1)(y-1)\leq 0$ as desired.$\blacksquare$