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I was just encountered an inequality in AoPs, Here it is:

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=4569&view=next,

that is:

If $a$, $b$ and $c$ are positive numbers, then we have inequality: $$\dfrac{(a^2+b^2)(b^2+c^2)(c^2+a^2)}{8}\ge\dfrac{(ab+bc+ca)^3}{27}$$

If the above ineq is succeed, it's easy to generalize the variables into real number.

Here I've got an terrible solution:

just expand(I think it is complicated to deal with this ineq using expand and easy to make mistake) and the inequality is equivalent to the following symmetric Ineq.

$$6a^2b^2c^2+27\sum_{sym} a^4b^2\ge4\sum_{sym} a^3b^3+24\sum_{sym} a^3b^2c,$$

then by the A-G Ineq, $$6a^2b^2c^2+\sum\limits_{sym} b^4c^2=\sum\limits_{sym}(a^4b^2+a^2b^2c^2)\ge2\sum\limits_{sym}a^3b^2c, $$ and Muirhead's theorem implies \begin{equation} \sum_{sym}a^4b^2\ge\sum_{sym}a^3b^2c,\quad \sum_{sym}a^4b^2\ge\sum_{sym}a^3b^3. \end{equation} finally the ineq can be proved by three appropriate coeffients multiplied to the previous three ineq and total them up.

Is there a concise proof without complicated calculating?(also it swept the buffalo way away). this is what I interested.

Michael Rozenberg
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Larry Eppes
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1 Answers1

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We'll prove that your inequality is true for all reals $a$, $b$ and $c$.

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

Since $\prod\limits_{cyc}(a^2+b^2)=\prod\limits_{cyc}(a^2+b^2+c^2-c^2)=-w^6+A(u,v^2)w^3+B(u,v^2)$,

we see that our inequality is equivalent to $f(w^3)\geq0$, where $f$ is a concave function,

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 even degree, it's enough to assume $b=c=1$,

which gives $27(a^2+1)^2\geq4(2a+1)^3$ or $(a-1)^2(27a^2+22a+23)\geq0$.

Done!

Michael Rozenberg
  • 203,855