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#1 h Jan 15, 2013, 9:49 PM • 1 Y

Y by Adventure10

1.Find all pairs of $(a,b)$ which are set of natural numbers and for that $\frac{1}{a}+\frac{1}{b}+\frac{a}{1+b}$ to be a natural number too.

2.Let $a,b,c$ are set of positiv real numbers and $a+b+c=3$ .Then prove $\sqrt{a}+\sqrt{b}+\sqrt{c}\geq ab+bc+ca$ .

3..Let $a,b,c$ are set of positiv real numbers and $a^2+b^2+c^2+(a+b+c)^2\leq4$ .Then prove $\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ca+1}{(c+a)^2}\geq3$ .

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#2 h Jan 16, 2013, 2:23 PM • 2 Y

Y by Adventure10, Mango247

2.Notice that: $2(ab+ac+bc)=(a+b+c)^2-(a^2+b^2+c^2)$
$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq ab+bc+ca$
$2(\sqrt{a}+\sqrt{b}+\sqrt{c})+a^2+b^2+c^2 \geq 9$
$a^2+2\sqrt{a}+b^2+2\sqrt{b}+c^2+2\sqrt{c}\geq 9$
Am-Gm inequality:
$\frac{a^2+\sqrt{a}+\sqrt{a}}{3}\geq a$
$\frac{b^2+\sqrt{b}+\sqrt{b}}{3}\geq b$
$\frac{c^2+\sqrt{c}+\sqrt{c}}{3}\geq c$
So, $\sqrt{a}+\sqrt{b}+\sqrt{c}\geq ab+bc+ca$

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#3 h Mar 27, 2025, 3:50 AM

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Kastriot wrote:

3..Let $a,b,c$ are set of positiv real numbers and $a^2+b^2+c^2+(a+b+c)^2\leq4$ .Then prove $\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ca+1}{(c+a)^2}\geq3$ .

https://artofproblemsolving.com/community/c5h404347p2254758

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