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The following inequality is exercise 1.8 from this book.

For any real $a,b,c$, prove the following $$(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \geq (ab+bc+ca)^3.$$

I've managed to prove this via brut-force and Muirhead's inequality (Very unsatisfying). However I'm having difficulty understanding the solution in the back of the book. I'm also interested if anyone else knows of any nice proofs of this inequality?

In particular, the solution at the back says the following (word for word):

If we can show that $\frac{27}{64}(a+b)^2(b+c)^2(c+a)^2\geq (ab+bc+ca)^2$, then the conclusion follows. Denote $S_1=a+b+c$, $S_2=ab+bc+ca$ and $S_3=abc$. We need to show that $27(S_1S_2-S_3)^2\geq 64S_2^3.$

It then goes onto prove the last inequality via cases. However I'm struggling to see how it is sufficient to prove either of those two inequalities.

We can rewrite $a^2+ab+b^2 = (\frac{a+b}{2})^2+\frac{3}{4}(a^2+b^2)$ and so by AM-GM it is sufficient to prove $$\sqrt{\frac{27}{64}(a+b)^2(b+c)^2(c+a)^2}\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)} \geq (ab+bc+ca)^3 .$$ However $$\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)} \geq (ab+bc+ca)^2 $$ is clearly wrong and so this does not justify the first claim in the solution. This does give me an idea of where the $\frac{27}{64}$ term comes from however. Other than this I'm a bit stuck and any help would be appreciated.

Edit: I realised straight after posting that $a^2+ab+b^2 \neq (\frac{a+b}{2})^2+\frac{3}{4}(a^2+b^2)$. So in fact my work is nonsense and can be ignored.

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    Are you sure that it isn't $\frac{27}{64}(a+b)^2(b+c)^2(c+a)^2\geq (ab+bc+ca)^3$ in the solution? – 2017-02-13
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    It must be a typo. Simply set $a=b=c =x$ (or think about homogeneity). – 2017-02-13
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    @Leon Sot Muirhead does not help for all real variables, which says that your reasoning is total wrong. See my solution. – 2017-02-13
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    @MichaelRozenberg Right, I forgot Muirhead is only for non-negative values. I'm looking at your answer now. – 2017-02-13

2 Answers 2

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It is a typo. In fact $27(S_1S_2-S_3)^2\geq 64S_2^3$ (which is correct) expands to $$ \frac{27}{64}(a+b)^2(b+c)^2(c+a)^2\geq (ab+bc+ca)^3 $$ with exponent $3$ instead of $2$ on the right-hand side.

Then the wanted inequality follows because $$ a^2 + ab + b^2 = \frac 34 (a+b)^2 + \frac 14 (a-b)^2 \ge \frac 34 (a+b)^2 $$

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    I think we need to prove that $27(a+b)^2(a+c)^2(b+c)^2\geq64(ab+ac+bc)^3$ for all real variables. – 2017-02-13
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    @MichaelRozenberg: As I understand it, the question was how to prove the inequality *using* the given hint from the book. – 2017-02-13
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    I think after this "hint" we obtain much harder inequality. By the way, the starting inequality and the second inequality are my old inequalities. – 2017-02-13
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    @MichaelRozenberg: I have only tried to answer the question which was asked here: *"I'm having difficulty understanding the solution in the back of the book ... However I'm struggling to see how it is sufficient to prove either of those two inequalities"* – 2017-02-13
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    @ Martin R You are right of course, but I think you understand now that the solution in the book is total wrong! – 2017-02-13
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We use the so called uvw method for proving inequality involving symmetric polynomials of $3$ variables.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.

Hence, our inequality it's $kw^6+A(u,v^2)w^3+B(u,v^2)\geq0$, where $A$ and $B$ polynomials such that $A(0,0)=B(0,0)=0$.

Let $\theta^3=1$, where $\theta\neq1$ and $(a,b,c)=(1,\theta,\theta^2)$.

Thus, $a^2+ab+b^2=a^2+ac+c^2=b^2+bc+c^2=u^2=v^2=0$, but $w^3\neq0$,

which says that $k=0$.

Thus, our inequality is a linear inequality of $w^3$,

which says that it's enough to prove our inequality for an extremal value of $w^3$,

which happens for equality case of two variables.

Since our inequality is homogeneous and even degree, we can assume $b=c=1$,

which gives $$3(a^2+a+1)^2\geq(2a+1)^3$$ or $$(a-1)^2(3a^2+4a+2)\geq0,$$ which is obvious.

Done!

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    Thank you for the alternate solution. The use of complex numbers is genius! (Are there any similar inequalities that use a similar sort of method?) One step I'm having difficulty seeing is the fact that $A(0,0)=B(0,0)=0$. Can you expand on that a little? – 2017-02-13
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    @Leon Sot Your inequality it's just $-u^3w^3+3u^2v^4-2v^6\geq0$. $A(u,v^2)=-u^3$ and $B(u,v^2)=3u^2v^4-2v^6$ and we see that indeed, $k=0$. We can understand this thing without expanding by using of complex numbers. – 2017-02-13
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    It seems to me that the strength of this approach is that we don't need to figure out the $A$ or $B$ explicitly. However, how do you know $A(0,0)=B(0,0)=0$ without figuring out the polynomials explicitly? – 2017-02-13
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    @Leon Sot $A$ is an homogeneous third degree polynomial of $a$, $b$, $c$ and polynomial of $u$ and $v^2$ only. Thus, $A(0,0)=0$. $B$ it's an homogeneous sixth degree polynomial of $a$, $b$, $c$ and it's polynomial of $u$ and $v^2$ only. Thus, $B(0,0)=0$. – 2017-02-13
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    Right! I get it now, thank you. – 2017-02-13
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    How does one prove a homogeneous polynomial 6'th degree of $(a,b,c)$ can always be written as $kw^6+A(u,v^2)w^3+B(u,v^2)$? Is it by using Newton's identity? – 2018-04-03
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    @Hans All symmetric polynomial of three variables we can write like polynomial of $u$, $v^2$ and $w^3$. For example, $a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)=9u^2-6v^2$. – 2018-04-03