Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Assuming $a,b,c \in (0, \infty)$, we need to prove that:

$$a+b+c \geqslant a b c+2 \quad \text{if} \quad ab+bc+ca=3$$

Can you give me an idea, please? This inequality seem to be known, but I didn't manage to solve it.

share|improve this question
    
I will go out on a limb and say that it can most likely be reduced to some mean inequality. (As in, an equality between means, not one that is not particularly nice. ;) ) –  tomasz Aug 10 '12 at 19:03
1  
@Iuli: how about using the fact that $\frac{ab+bc+ca}{3}=1 \ge (abc)^\frac{2}{3}$? –  Chris's sis Aug 10 '12 at 19:09
    
I think the clever solution will probably start with the observation that (a+1)(b+1)(c+1)=abc+a+b+c+4 or something of the sort... Not sure where to go next. It's been quite some time since I'd been solving inequalities. –  tomasz Aug 10 '12 at 19:12
    
@tomasz: first time I had in my mind this point, but it can be solved easier. –  Chris's sis Aug 10 '12 at 19:13
    
Some thoughts, and maybe someone could rewrite it as an answer. $a,b,c\in(0,\infty)$. Let $\alpha,\beta, \gamma \in(o,\frac{\pi}{2})$, and: $a=\tan\alpha, \ b=\tan\beta, \ c=\tan\gamma$. We can prove that, $$\tan(\alpha + \beta + \gamma)=\frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1 - \tan\alpha\tan\beta-\tan\beta\tan\gamma-\tan\gamma\tan\alpha}$$ The problem is with $\tan(\alpha + \beta + \gamma)$ –  Salech Alhasov Aug 10 '12 at 19:36

4 Answers 4

up vote 16 down vote accepted

Let's start by applying AM-GM:

$$\frac{ab+bc+ca}{3}=1 \ge (abc)^\frac{2}{3}$$ $$ 1 \ge abc$$ Further we get that $$3\ge abc+2 \tag1$$ On the other hand we have
$$ (a+b+c)^2 \ge 3(ab+bc+ac)=9 $$ $$a+b+c \ge 3 \tag2$$

From $(1)$ and $(2)$ we get the required inequality $$a+b+c \ge abc+2$$

Q.E.D.

share|improve this answer
1  
+1 ...... wow!! –  Santosh Linkha Aug 10 '12 at 22:11

Solve for $c$ from the constraint: $$ c = \frac{3- a b}{a+b} $$ and substitute back into the inequality: $$ a + b + \frac{3-a b}{a+b} \geqslant 2 + \frac{a b(3-a b)}{a+b} \implies \frac{3 + a^2 b^2 + a^2 + b^2 -2 a b -2a -2b+3}{a+b} \geqslant 0 $$ Multiply both sides by $(a+b)$: $$ 3 + a^2 b^2 + a^2 + b^2 -2 a b -2a -2b+3 \geqslant 0 $$ The left-hand-side can be reduced to the shifted sum of squares: $$ (a b-2)^2 + (a+b-1)^2 - 2 \geqslant 0 $$ It is easy to see that the left-hand-side attains its minimum at $a=b=1$, where the inequality is saturated.

share|improve this answer

For $x,y,z \geq 0 $, $ f(x,y,z) = x+y+z $ given that, $xy+yz+xz =3 = \phi(x, y, z)$

$$ \nabla f = \lambda \nabla \phi $$

$$ 1 = \lambda (y+z) \hspace {2 cm} (1) $$ $$ 1 = \lambda (x+z) \hspace {2 cm} (2) $$ $$ 1 = \lambda (x+y) \hspace {2 cm} (3) $$ $$ xy+yz+xz =3 \hspace{2 cm} (4)$$ Solveing $(1), (2), (3), \text{ and } (4)$ we get, $x=y=z=1$, so the minimum value of $f(x, y, z) = x+y+z$ under the constraint $xy+yz+xz = 3$ is $ 1 + 1 + 1 = 3$

Again $g(x, y, z) = xyz + 2$ $$ \nabla g = \lambda \nabla \phi $$

$$ yz = \lambda (y+z) \hspace {2 cm} (5) $$ $$ xz = \lambda (x+z) \hspace {2 cm} (6) $$ $$ xy = \lambda (x+y) \hspace {2 cm} (7) $$ $$ xy+yz+xz =3 \hspace{2 cm} (4)$$ Solving these we get $x=y=z=1$, the maximum value of $g(x, y, z) = 1\cdot 1 \cdot 1 + 2 = 3 $
Since, $ \text{ min }( f) = \text{ max}(g) $, we have $a+b+c \geq abc+2$

I hope there is a better method!!

share|improve this answer
    
Nice. Equation (2) should have $\lambda(x+z)$ however :) –  Arkamis Aug 10 '12 at 19:08
1  
Lagrange multipliers ... thanks :) –  Santosh Linkha Aug 10 '12 at 19:09

It's easy to see that $(x+y+z)^2 \geq 3(xy+yz+zx)$.Denoting $S=a+b+c$ we cand deduce from this that $S\geq 3$.

Plugging in $x=ab$, $y=bc$, $z=ca$ we get $9\geq 3abc(a+b+c)$, so $abc \leq \frac{3}{a+b+c}$.So it's enough to prove that $a+b+c \geq \frac{3}{a+b+c}+2$.

The last inequality is equivalent to $S^2-2S-3 \geq 0$ which is true since $S\geq 3$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.