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.
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$\begingroup$ 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. ;) ) $\endgroup$– tomaszAug 10, 2012 at 19:03
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1$\begingroup$ @Iuli: how about using the fact that $\frac{ab+bc+ca}{3}=1 \ge (abc)^\frac{2}{3}$? $\endgroup$– user 1591719Aug 10, 2012 at 19:09
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$\begingroup$ 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. $\endgroup$– tomaszAug 10, 2012 at 19:12
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$\begingroup$ @tomasz: first time I had in my mind this point, but it can be solved easier. $\endgroup$– user 1591719Aug 10, 2012 at 19:13
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1$\begingroup$ 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 $\tan(\alpha + \beta + \gamma)$ $\endgroup$– Salech AlhasovAug 10, 2012 at 19:36
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4 Answers
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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.
answered Aug 10, 2012 at 19:15
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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.
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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!!
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$\begingroup$ Nice. Equation (2) should have $\lambda(x+z)$ however :) $\endgroup$– EmilyAug 10, 2012 at 19:08
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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 $S\geq 3$.
