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Home » Olympiad Section » Inequalities » Inequalities Unsolved Problems
 
a/(a^2+5bc)
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dduclam
Riemann Hypothesis
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#1
a/(a^2+5bc)
For a,b,c be sidelengths of a triangle, show that
\frac a{a^2+5bc}+\frac b{b^2+5ca}+\frac c{c^2+5ab}\ge\frac{a+b+c}{2(ab+bc+ca)}
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Duong Duc Lam

PostPosted: Tue Nov 03, 2009 6:15 pm  Back to top 
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can_hang2007
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#2
Re: a/(a^2+5bc)
dduclam wrote:
For a,b,c be sidelengths of a triangle, show that
\frac a{a^2 + 5bc} + \frac b{b^2 + 5ca} + \frac c{c^2 + 5ab}\ge\frac {a + b + c}{2(ab + bc + ca)}

After using the Cauchy-Schwarz Inequality, we see that it suffices to show that the following inequality holds
2(a+b+c)(ab+bc+ca) \ge a^3+b^3+c^3+15abc,
which is true and it is not hard to prove.
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PostPosted: Tue Nov 03, 2009 8:22 pm  Back to top 
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mestav
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#3
Can_hang2007 please prove your inequality. I replaced a => x+y, b=>y+z, c=>x+z and multiplied and simplied. I have (3,0,0)+(1,1,1) >= 2 (2,1,0) at your inequality. I think it is false

PostPosted: Wed Nov 04, 2009 10:39 am  Back to top 
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can_hang2007
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mestav wrote:
Can_hang2007 please prove your inequality. I replaced a => x+y, b=>y+z, c=>x+z and multiplied and simplied. I have (3,0,0)+(1,1,1) >= 2 (2,1,0) at your inequality. I think it is false

Dear friend,
The inequality
\sum_{sym} x^3 + \sum_{sym} xyz \ge 2\sum_{sym} x^2y
is true and it is known as Schur's Inequality.

We can prove it by writing it as
x(x-y)(x-z)+y(y-z)(y-x)+z(z-x)(z-y) \ge 0,
or
(x-y)^2(x+y-z)+z(x-z)(y-z) \ge 0.
Now, by assuming z=\min\{x,y,z\}, we deduce the result.
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PostPosted: Wed Nov 04, 2009 4:56 pm  Back to top 
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dduclam
Riemann Hypothesis
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#5
Re: a/(a^2+5bc)
can_hang2007 wrote:
dduclam wrote:
For a,b,c be sidelengths of a triangle, show that
\frac a{a^2 + 5bc} + \frac b{b^2 + 5ca} + \frac c{c^2 + 5ab}\ge\frac {a + b + c}{2(ab + bc + ca)}

After using the Cauchy-Schwarz Inequality, we see that it suffices to show that the following inequality holds
2(a + b + c)(ab + bc + ca) \ge a^3 + b^3 + c^3 + 15abc,
which is true and it is not hard to prove.


You are right, Can. The final inequality equivalent to (3c-a-b)(a-b)^2+(a+b-c)(a-c)(b-c)\ge0, which obviously if we assume that c=\max\{a,b,c\}.

Now, what do you think about the following inequality?
If a,b,c be sidelenghths of an acute triangle, then
\frac a{a^2 + 3bc} + \frac b{b^2 + 3ca} + \frac c{c^2 + 3ab}\ge\frac 1{2}\left(\frac1{a+b}+\frac1{b+c}+\frac1{c+a}\right)
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Duong Duc Lam

PostPosted: Wed Nov 04, 2009 5:24 pm  Back to top 
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