# Inequality $a+b+c \geqslant abc +2$

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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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
@Iuli: how about using the fact that $\frac{ab+bc+ca}{3}=1 \ge (abc)^\frac{2}{3}$? – Chris's sis the artist 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 the artist 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

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.

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+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.

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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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Nice. Equation (2) should have $\lambda(x+z)$ however :) – Arkamis Aug 10 '12 at 19:08
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 $S\geq 3$.

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