abc=1

Maverick

Let $a,b,c$ be three positive real numbers such that $abc=1$ . Prove that: $1+\frac{3}{a+b+c}\ge{\frac{6}{ab+bc+ca}} .$

**Posted:** Sat Jul 31, 2004 8:32 am

fuzzylogic · Yang-Mills Theory Offline Joined: 24 Mar 2004 Posts: 719

$\displaystyle 1+\frac{3}{a+b+c}\geq \frac{6}{ab+bc+ca}$ is equivalent to

$\displaystyle \frac{ab+bc+ca}{3} + \frac{ab+bc+ca}{a+b+c} \geq 2$ .

By AM-GM, it suffices to prove $\displaystyle \frac{ab+bc+ca}{3} \cdot \frac{ab+bc+ca}{a+b+c} \geq 1$ .

That is, $\displaystyle (ab+bc+ca)^2 \geq 3(a+b+c)$ .

Let $x=1/a, y=1/b, z=1/c$ , then $xyz=1$ and it reduces to prove $(x+y+z)^2\geq 3(xy+yz+zx)$ which is obvious.

**Posted:** Sat Jul 31, 2004 9:42 am

asdrojas · P versus NP Offline Joined: 06 Aug 2007 Posts: 33

$1 + \frac {3}{a + b + c}\ge{\frac {6}{ab + bc + ca}}$ is equivalent to $(3+a + b + c)(ab + bc + ca)\ge 6(a + b + c)$ .

Now lets consider two cases. First $ab + bc + ca\ge a + b + c$ and then $ab + bc + ca\le a + b + c$ .

If $ab + bc + ca\ge a + b + c$ then using AM-GM $(3+a + b + c)(ab + bc + ca)\ge (3+3)(ab + bc + ca))\ge 6(ab + bc + ca)\ge 6(a + b + c)$ .

If $ab + bc + ca\le a + b + c$ then let $x= a + b + c$ .

$(3+a + b + c)(ab + bc + ca)\ge (3+a + b + c)(a + b + c) = (3+x)x$ . Now we net that $(3+x)x\ge 6x$ but this equivalent to $x(x-3)\ge 0$ and this is true becouse $a+b+c\ge 3$ by AM-GM.

**Posted:** Fri Jan 09, 2009 4:37 pm

phuocphu

**Posted:** Sat Jan 10, 2009 5:51 pm

Mathias_DK

**Posted:** Sun Jan 11, 2009 1:48 pm

CCMath1 · Poincare Conjecture Offline Joined: 29 Jun 2009 Posts: 127

first ,we use ${abc = 1}$ ,so the inequality ${\Leftrightarrow}$ ${\frac {6}{\frac {1}{a} + \frac {1}{b} + \frac {1}{c}} < = 1 + \frac {3}{a + b + c}}$ ;we know that : ${\frac {3}{\frac {1}{a} + \frac {1}{b} + \frac {1}{c}} \le \frac {a + b + c}{3}}$ so we only need to prove that : ${\frac {2}{3}(a+b+c) \le 1 + \frac {3}{a + b + c}}$ _______[#];let ${a + b + c = x}$ by ${a + b + c \le 3*(abc)^{\frac {1}{3}}}$ ,we know ${x \in (0,3] }$ ; so [#] ${\Longleftrightarrow}$ ${\frac {2}{3}x^2 - x - 3 > = 0}$ ; then consider the function: ${\phi (x) = \frac {2}{3}(x - \frac {3}{4})^2 - \frac {27}{8}}$ ; we know that : ${\phi (x) \ge 0}$ on ${x \in (0,3]}$ . then # is true ,so we have completed!

**Posted:** Fri Nov 06, 2009 2:22 am

Toan_VN_LC

**Posted:** Fri Nov 06, 2009 2:53 am