Inequality question.

Question

Let $a,b,c>0$ with $a+b+c=1$. Show that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq 3 + 2\cdot\frac{\left(a^3 + b^3 + c^3\right)}{abc}$$

Ohhhkk. So first off,

\begin{align} a^3 + b^3+ c^3 & =a^3 + b^3+ c^3- 3abc +3abc\\ & =\ (a+b+c)(a^2+b^2+c^2-(ab+bc+ca))+3abc\\ & = \ (1-3(ab+bc+ca)) + 3abc \\ \end{align}

Using this the inequality becomes, $$7 \cdot \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \leq 9+ \frac{2}{abc}$$

How do i proceed from here? Was this the right approach? Is there a better one?

Answer 1

I think the following is a smooth enough.

We need to prove that: $$(a+b+c)(ab+ac+bc)\leq3abc+2(a^3+b^3+c^2)$$ or $$\sum_{cyc}(2a^3-a^2b-a^2c)\geq0$$ or $$\sum_{cyc}(a-b)^2(a+b)\geq0,$$ which is obvious.

Your second inequality it's indeed just the first: $$7\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\leq9+\frac{2}{abc}$$ it's $$7(a+b+c)(ab+ac+bc)\leq9abc+2(a+b+c)^3$$ or $$\sum_{cyc}(3abc+2a^3+6a^2b+6a^2c+4abc-7a^2b-7a^2c-7abc)\geq0$$ or $$\sum_{cyc}(2a^3-a^2b-a^2c)\geq0$$ or $$\sum_{cyc}(a^3-a^2b-ab^2+b^3)\geq0$$ or $$\sum_{cyc}(a-b)^2(a+b)\geq0.$$

Answer 2

Hint: Just expand and you're done.

(As per Michael's first step) Expanding out the denominator and homogenizing, we WTS

$$ (a+b+c)(ab+bc+ca) \leq 3 abc + 2(a^3+b^3+c^3)$$

Expanding terms and cancelling $abc$, we WTS

$$ a^2b + b^2 a + c^2 b + b^2 a + c^2 b + a^2 c \leq 2 (a^3 + b^3 + c^3).$$

This should be obvious / well known e.g.

$a^2b + b^2 a + c^2 b \leq a^3 + b^3 + c^3$ and $b^2 a + c^2 b + a^2 c \leq a^3 + b^3 + c^3$.

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Unsurprisingly, since you applied an identity (as opposed to an inequality), we can follow the same steps of homogenizing using your approach.

We WTS

$$ 7 (ab+bc+ca)(a+b+c) \leq 9 abc + 2(a+b+c)^3$$

Expanding, and cancelling common terms, this becomes:

$$ a^2b + b^2 a + c^2 b + b^2 a + c^2 b + a^2 c \leq 2 (a^3 + b^3 + c^3).$$