I sorry I have just see that it wrong
It must be
\[\frac{a}{3+2\sqrt{3}+b^2}+\frac{b}{3+2\sqrt{3}+c^2}+\frac{c}{3+2\sqrt{3}+a^2}\ge \frac{3}{4+2\sqrt{3}}\]
( this is Vasc's conjecture)
Search found 7 matches
- Sat Aug 04, 2012 1:29 pm
- Forum: Algebra
- Topic: hard inequality 002
- Replies: 5
- Views: 3869
- Sat Aug 04, 2012 9:44 am
- Forum: Algebra
- Topic: hard inequality 002
- Replies: 5
- Views: 3869
Re: hard inequality 002
$a=0,47; b=1,01; c=1,52$
Then
$$\sum ab^2 =3,15$$
Then
$$\sum ab^2 =3,15$$
- Fri Aug 03, 2012 12:32 pm
- Forum: Algebra
- Topic: hard inequality 002
- Replies: 5
- Views: 3869
Re: hard inequality 002
I sorry. But this is wrong !SANZEED wrote:$\sum_{cyclic}ab^{2}\leq 3$.
- Thu Aug 02, 2012 12:23 pm
- Forum: Algebra
- Topic: inequality 005
- Replies: 0
- Views: 1472
inequality 005
Let $a,b,c \ge 0$ satisfy $a+b+c=3$ Show that
\[a^2+b^2+c^2+\sqrt{3}.abc\sqrt{\sqrt{a}+\sqrt{b}+\sqrt{c}} \ge 6\]
\[a^2+b^2+c^2+\sqrt{3}.abc\sqrt{\sqrt{a}+\sqrt{b}+\sqrt{c}} \ge 6\]
- Thu Aug 02, 2012 12:22 pm
- Forum: Algebra
- Topic: hard inequality 002
- Replies: 5
- Views: 3869
hard inequality 002
Let $a,b.c \ge 0$ satisfy $a+b+c=3$ Show that
\[\frac{a}{6+\sqrt{3}+b^2}+\frac{b}{6+\sqrt{3}+c^2}+\frac{c}{6+\sqrt{3}+a^2}\ge \frac{3}{7+\sqrt{3}}\]
\[\frac{a}{6+\sqrt{3}+b^2}+\frac{b}{6+\sqrt{3}+c^2}+\frac{c}{6+\sqrt{3}+a^2}\ge \frac{3}{7+\sqrt{3}}\]
- Fri Jul 13, 2012 9:45 am
- Forum: Algebra
- Topic: Solve equation
- Replies: 0
- Views: 1535
Solve equation
$a) \sqrt{\frac{x^3}{3-4x}}-\frac{1}{2\sqrt{x}}=\sqrt{x}$
$b) \sqrt{x+1}+6\sqrt{9-x^2}+6\sqrt{(x+1)(9-x^2)}=38+10x-2x^2-x^3$
$c) x^3=\sqrt{x^4-1}+\sqrt{x^2-1}$
\[x\in \mathbb{R}\]
I want one detail solution !
$b) \sqrt{x+1}+6\sqrt{9-x^2}+6\sqrt{(x+1)(9-x^2)}=38+10x-2x^2-x^3$
$c) x^3=\sqrt{x^4-1}+\sqrt{x^2-1}$
\[x\in \mathbb{R}\]
I want one detail solution !
- Thu Jul 12, 2012 10:17 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2012: Day 1 Problem 2
- Replies: 12
- Views: 19083
Re: IMO 2012: Day 1 Problem 2
By inequality Holder we have $$(1+1)^1(1+a_2)^2(1+a_3)^3...(1+a_n)^n \ge (1+a_2.a_3...a_n)^{\frac{n(n+1)}{2}}$$ $$=2^{\frac{n(n+1)}{2}} = \left( 2^{\frac{(n+1)}{2}}\right)^n \ge ( 2\ln2.n )^n=n^n(2\ln2)^n > 2n^n, (n \ge 3)$$ Note $$f(x)=2^{\frac{x+1}{2}}-2\ln2.x \ge 0, \forall x \ge 3$$ So $$(1+a_2)...