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Let $a,b,c\ge0$. Prove that:

$$\frac ab+\frac bc+\frac ca+\frac{3\sqrt[3]{abc}}{a+b+c}\ge4$$

This is a problem from Samin Riasat's Olympiad Inequalities worksheet. The worksheet gives a hint as to prove and use $$\frac ab+\frac bc+\frac ca\geq\frac{a+b+c}{\sqrt[3]{abc}}$$ I've done it, substituted it in the original inequality and now I have to prove:

$$\frac{9\mathrm{AM}^2+3\mathrm{GM}^2}{3\mathrm{AM}\cdot\mathrm{GM}}\geq4$$

where $\mathrm{AM}$ and $\mathrm{GM}$ are the arithmetic and geometric means of $a$, $b$ and $c$. How can I finish the problem?

ultralegend5385
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PabloGamerX
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  • You can use uvw (also known as pqr) or just simplify and use Muirhead or something. – D S Mar 29 '24 at 12:03
  • FYI. A stronger inequality is https://math.stackexchange.com/questions/4874438/elegant-proof-of-frac25-left-fracxy-fracyz-fraczx-right-f – River Li Mar 29 '24 at 12:49
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    Then let $x = \frac{a+b+c}{\sqrt[3]{abc}} \ge 3$ (by AM-GM), you want to prove $x + 3/x \ge 4$ or $(x - 1)(x-3)/x \ge 0$ which is true. – River Li Mar 29 '24 at 13:33
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    Check https://math.stackexchange.com/questions/189143/proving-inequality-fracab-fracbc-fracca-frac3-sqrt3abca?noredirect=1 or https://math.stackexchange.com/questions/3703357/inequality-involving-am-gm-but-its-wierd?noredirect=1 or even https://math.stackexchange.com/questions/185604/prove-inequality-fracab-fracbc-fracca-frac9-sqrt3abcab?noredirect=1 or https://math.stackexchange.com/questions/3764022/strange-cube-root-offense-in-an-inequality?noredirect=1 – Macavity Mar 29 '24 at 17:32

2 Answers2

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I think I have a solution. Take the expression with AM and GM, it simplifies to $3\frac{AM}{GM}+\frac{GM}{AM}$. Now, since $AM\ge GM$, there exists $k \ge 1$ such that $AM=k\cdot GM$. Substituting in the inequality, we get $$3k+\frac1k \ge 4$$ Taking the derivative, we get $3-1/k^2$. For $k \ge 1$, the derivative is positive, so the minimum value of the expression for $k \ge 1$ is the value for $k=1$, which is $4$, and it finishes the problem.

D S
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PabloGamerX
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First, let's see this:- $$a^3b^3c^3=(a^2c)(b^2a)(c^2b)=(abc)^3$$ So we can say from the second expression, using AM-GM inequality that:-

$$\frac{a^2c+b^2a+c^2b}{3}≥\sqrt[3]{(a^2c)(b^2a)(c^2b)}$$ Or $$a^2c+b^2a+c^2b≥3\sqrt[3]{(abc)^3}$$ Or $$\frac{a^2c+b^2a+c^2b}{abc}≥3...(i)$$

Again, by AM-GM inequality we know that:- $$\frac{a+b+c}{3}≥\sqrt[3]{abc}$$ Or $$\frac{a+b+c}{3\sqrt[3]{abc}}≥1$$ Or (By doing reciprocal) $$\frac{3\sqrt[3]{abc}}{a+b+c}≤1$$ or $$1≥\frac{3\sqrt[3]{abc}}{a+b+c}...(ii)$$

Adding (i) and (ii), we get $$\frac{a^2c+b^2a+c^2b}{abc} + 1 ≥ 3 + \frac{3\sqrt[3]{abc}}{a+b+c}$$ Or $$\frac{a^2c+b^2a+c^2b}{abc}-\frac{3\sqrt[3]{abc}}{a+b+c}≥2$$ Adding $\frac{6\sqrt[3]{abc}}{a+b+c}$ on both sides $$\frac{a^2c+b^2a+c^2b}{abc}+\frac{3\sqrt[3]{abc}}{a+b+c}≥2+\frac{6\sqrt[3]{abc}}{a+b+c}$$

But from $(ii)$, we know $$2≥\frac{6\sqrt[3]{abc}}{a+b+c}$$ Or $$4≥2+\frac{6\sqrt[3]{abc}}{a+b+c}$$

Hence $$\frac{a^2c+b^2a+c^2b}{abc}+\frac{3\sqrt[3]{abc}}{a+b+c}≥4$$

I didn't use your hint somehow

Gwen
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    If $P\geq Q$ and $4\geqslant Q$ that does not mean $P\geqslant 4$. Your last step is faulty. – Macavity Mar 29 '24 at 17:18
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    @Macavity The maximum value of Q is 4, while P is more than or equal to any value of Q, be it 4 or less. Can you give an example where this doesn't happen? Reply edit:- Okay I got it. Q is not compulsorily 4. It can be less. Can you please provide some help? – Gwen Mar 29 '24 at 17:20
  • This has been done multiple times at this site IIRC. Will provide some links above. – Macavity Mar 29 '24 at 17:30