Find all functions $f: [0, \infty ) \rightarrow [0, \infty )$ which satisfies the following inequality:
$\sqrt{3} f(2x)+5 f(2y) \le 2 f(\sqrt{3} x+5y), \, \forall x,y \ge 0$.
Search found 53 matches
Equation
Solve the equation:
$\displaystyle \frac{1}{1+\sqrt{2} \sin x}+\frac{1}{1+ \sqrt{2} \cos x}=(\sqrt{2})^{1+x} , \, x \in \left( -\frac{\pi}{4}, \frac{\pi}{4} \right).$
$\displaystyle \frac{1}{1+\sqrt{2} \sin x}+\frac{1}{1+ \sqrt{2} \cos x}=(\sqrt{2})^{1+x} , \, x \in \left( -\frac{\pi}{4}, \frac{\pi}{4} \right).$
Sum...
Let m,n be two natural numbers. Prove that:
$\displaystyle \sum\limits_{i=-m}^{n} (-1)^i \frac{(2m+i)! (2n-i)!}{((m+i)!)^2((n-i)!)^2}=1$
$\displaystyle \sum\limits_{i=-m}^{n} (-1)^i \frac{(2m+i)! (2n-i)!}{((m+i)!)^2((n-i)!)^2}=1$
Ineq 3
Let a,b,c,d>0 with $ab+bc+cd+da \le 8$. Prove that:
$\frac{a^2+b^2}{(a+b)^4}+\frac{b^2+c^2}{(b+c)^4}+\frac{c^2+d^2}{(c+d)^4}+\frac{d^2+a^2}{(d+a)^4} \le \frac{1}{abcd}$
$\frac{a^2+b^2}{(a+b)^4}+\frac{b^2+c^2}{(b+c)^4}+\frac{c^2+d^2}{(c+d)^4}+\frac{d^2+a^2}{(d+a)^4} \le \frac{1}{abcd}$
- Tue Apr 16, 2013 1:42 am
- Forum: Number Theory
- Topic: Natural numbers
- Replies: 4
- Views: 3369
Re: Natural numbers
Well...I think a and b must be prime...but I must prove that (if it is so)
- Mon Apr 15, 2013 12:11 pm
- Forum: Number Theory
- Topic: Natural numbers
- Replies: 4
- Views: 3369
Re: Natural numbers
Yes......!
Ineq 2
Prove that, if $x,y,z \in [1, \frac{4}{3}]$, then:
a)$x \sqrt{4-3z}+y \sqrt{4-3x}+z \sqrt{4-3y} \le 4$;
b)$xy \sqrt{4-3z}+xz \sqrt{4-3y}+yz \sqrt{4-3x} \le x^2+y^2+z^2$.
a)$x \sqrt{4-3z}+y \sqrt{4-3x}+z \sqrt{4-3y} \le 4$;
b)$xy \sqrt{4-3z}+xz \sqrt{4-3y}+yz \sqrt{4-3x} \le x^2+y^2+z^2$.
Re: Ineq
Thank you!
Ineq
Let a,b,c>0 be three real numbers. Prove that:
$\sum\limits_{cyc} \frac{a(b+c)}{\sqrt{(a^2+b^2)(a^2+c^2)}} \le 3$
$\sum\limits_{cyc} \frac{a(b+c)}{\sqrt{(a^2+b^2)(a^2+c^2)}} \le 3$
- Mon Apr 08, 2013 2:38 pm
- Forum: Number Theory
- Topic: Natural numbers
- Replies: 4
- Views: 3369
Natural numbers
Find all pairs of natural numbers a and b such that:
$1+ \sigma (a) + \sigma (b) +\sigma (ab) =(a+2)(b+2)$
and
$\text{lcd}(a,b)+\gcd(a,b)+ \sigma(a)=2066$
where $\sigma (x)$=the sum of (natural) divisors of x.
$1+ \sigma (a) + \sigma (b) +\sigma (ab) =(a+2)(b+2)$
and
$\text{lcd}(a,b)+\gcd(a,b)+ \sigma(a)=2066$
where $\sigma (x)$=the sum of (natural) divisors of x.